Electric Field Due To Point Charges - Physics Problems

00:59:58
https://www.youtube.com/watch?v=V9RLc9EX1so

摘要

TLDRDans cette vidéo, le discours détaillé se concentre sur les champs électriques, leur calcul et leur influence sur des charges test dans leur environnement. Le champ électrique est essentiel pour comprendre la direction et l'intensité de la force agissant sur les charges. Il est représenté par la formule E = F/q, avec F étant la force en newtons et q la charge en coulombs. La vidéo explique comment les champs diffèrent pour les charges positives et négatives : un champ électrique exercera une force dans le même sens sur une charge positive mais dans le sens opposé sur une charge négative. Les aspects pratiques incluent des calculs de champs à des points spécifiques par rapport à des charges ponctuelles et des discussions sur la manière dont les champs changent avec l'augmentation ou la diminution de la distance ou de la magnitude de la charge. La vidéo contient enfin des problèmes pratiques qui illustrent l'application des concepts abordés pour des calculs précis et la détermination des directions vectorielles du champ électrique.

心得

  • 🔋 Le champ électrique se calcule par E = F/q.
  • ➕ Un champ pour une charge positive est extérieur.
  • ➖ Un champ pour une charge négative est intérieur.
  • 🔄 Un champ électrique est un vecteur comme la force.
  • ⚡ Les charges créent leur propre champ électrique.
  • 🧲 La direction du champ dépend de la polarité de la charge.
  • 📏 Un champ change avec le carré de la distance.
  • 🔍 Analyse de champs autour de charges ponctuelles.
  • 🧮 Utilisation des lois de Coulomb pour calculer les forces.
  • 📚 Problèmes pratiques démontrant l'application des théories des champs électriques.

时间轴

  • 00:00:00 - 00:05:00

    Le concept d'un champ électrique est introduit avec l'équation qui le définit comme étant la force électrique divisée par la charge. Le champ électrique est un vecteur qui indique la direction dans laquelle une charge se déplacera ; pour une charge positive, il est dans la même direction que le champ, et pour une charge négative, il est dans la direction opposée.

  • 00:05:00 - 00:10:00

    Les charges peuvent créer leur propre champ électrique : un champ créé par une charge positive rayonne de cette charge, tandis qu'un champ créé par une charge négative converge vers elle. Une équation est présentée pour calculer le champ électrique dû à une charge ponctuelle donnée. Un exemple est utilisé pour illustrer comment la petite charge d'essai est théoriquement placée pour mesurer le champ électrique sans influencer lui-même le champ.

  • 00:10:00 - 00:15:00

    Le calcul du champ électrique pour une charge ponctuelle est revisité avec la simplification en annulant la charge d'essai dans l'équation, offrant une formule simplifiée : E = kQ/r². Des calculs d'un champ électrique avec différentes unités de charge (micro, nano, milli coulombs) sont expliqués. On explique comment dessiner des vecteurs de champ électrique autour d'une charge dans un espace autour de celle-ci.

  • 00:15:00 - 00:20:00

    La direction du champ électrique produite par des charges positives et négatives est discutée avec des exemples de points autour des charges positives et négatives. Un problème est donné sur la détermination de la direction du champ électrique à un point donné, illustrant comment une charge négative est influencée dans un champ dirigé vers le sud quand la force est vers le nord.

  • 00:20:00 - 00:25:00

    On continue avec un problème concret où une charge positive doit être suspendue entre deux plaques métalliques, déterminant la masse nécessaire pour que la force électrique égale la force de gravité afin de rester suspendu. L'idée est que deux plaques parallèles créent un champ électrique uniforme.

  • 00:25:00 - 00:30:00

    Un scénario est exploré où un électron dans un champ électrique uniformément orienté à l'ouest est étudié pour trouver l'accélération et la vitesse à laquelle il serait éjecté. Les lois du mouvement de Newton sont utilisées pour dériver l'équation nécessaire utilisant F = ma pour un calcul de champ électrique.

  • 00:30:00 - 00:35:00

    Les conséquences sur le champ électrique lorsqu'une charge double ou que sa distance change sont traitées. On montre que doubler la charge double le champ électrique, tandis que doubler la distance le réduit d'un quart. Réduire la distance par un facteur de trois augmente le champ par neuf.

  • 00:35:00 - 00:40:00

    Une série d'exemples est donnée pour illustrer comment des champs électriques créés par deux charges identiques s'annulent exactement au point milieu entre elles, à cause de leur symétrie, sauf si la magnitude change. D'autres exemples montrent ce qui arrive quand l'une des charges double : le point d'annulation du champ se déplace.

  • 00:40:00 - 00:45:00

    On aborde les concepts de superposition des champs électriques et comment déterminer où ces champs se neutralisent. Des exemples visuels aident à imaginer comment les directions des vecteurs de champ électrique changent avec la position et la charge.

  • 00:45:00 - 00:50:00

    La méthode pour résoudre des problèmes de champ électrique en considérant les positions relatives des charges, conduisant à des vecteurs de direction opposés égalisant des côtés opposés, est exposée. Des explications détaillées montrent comment approcher les problèmes de physique en utilisant la théorie mathématique.

  • 00:50:00 - 00:59:58

    Le comportement d'un électron dans un champ électrique est étudié, montrant ses caractéristiques et études de cas pour dériver des formules pratiques qui résolvent des problématiques basées sur le champ électrique, utilisant encore F = ma et les formules d'accélération.

显示更多

思维导图

Mind Map

常见问题

  • Comment calcule-t-on un champ électrique?

    Le champ électrique est déterminé par la formule F/q, où F est la force électrique en newtons et q est la charge en coulombs.

  • Quelle est la direction du champ électrique pour des charges positives et négatives?

    Pour une charge positive, le champ électrique se dirige vers l'extérieur de la charge, tandis que pour une charge négative, il se dirige vers la charge.

  • Quand le champ électrique net devient-il zéro?

    Au point milieu entre des charges égales, les champs électriques des deux charges s'annulent mutuellement.

  • Quelle est la relation entre la direction de la force et celle du champ électrique sur des charges positives et négatives?

    La force électrique et le champ électrique sont dans la même direction pour une charge positive et dans des directions opposées pour une charge négative.

  • Les charges peuvent-elles créer des champs électriques?

    Oui, les charges peuvent créer leurs propres champs électriques selon leur polarité.

  • Comment le champ électrique change-t-il avec la magnitude de la charge et la distance?

    Le champ électrique change proportionnellement avec la charge source et inversement avec le carré de la distance.

  • Quelles sont les unités du champ électrique?

    Les unités de champ électrique sont newtons par coulomb (N/C).

  • Comment deux charges identiques affectent-elles le champ électrique à leur point milieu?

    Deux charges identiques créent un champ électrique nul au point milieu entre elles lorsque leurs magnitudes sont les mêmes et qu'elles sont équidistantes.

  • Comment un champ électrique uniforme affecte-t-il une charge?

    Dans un champ électrique uniforme, la force sur une charge est constante et dépend de la charge elle-même et de la force du champ.

  • Comment un champ électrique uniforme est-il créé?

    Des plaques parallèles peuvent créer un champ électrique uniforme, influençant les charges placées entre elles.

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  • 00:00:01
    in this video we're going to focus on
  • 00:00:03
    electric fields
  • 00:00:05
    let's begin our discussion
  • 00:00:07
    with the formula
  • 00:00:09
    that will help us to calculate the
  • 00:00:10
    electric field
  • 00:00:11
    the electric field is equal to the
  • 00:00:13
    electric force
  • 00:00:14
    that is acting on
  • 00:00:17
    a
  • 00:00:18
    tiny test charge
  • 00:00:20
    divided by
  • 00:00:21
    the magnitude of that test charge
  • 00:00:25
    so it's f over q
  • 00:00:27
    the force is measured in newtons
  • 00:00:30
    the charge q is measured in coulombs
  • 00:00:34
    so electric field
  • 00:00:36
    has the units newtons per coulomb
  • 00:00:42
    now the electric field is a vector
  • 00:00:44
    much like force is a vector
  • 00:00:47
    but the good thing about the electric
  • 00:00:48
    field is
  • 00:00:49
    it tells you
  • 00:00:51
    how the electric force is going to be
  • 00:00:53
    acting
  • 00:00:54
    on a positive or negative
  • 00:00:56
    charge
  • 00:00:57
    so let's talk about positive charges
  • 00:00:59
    first
  • 00:01:01
    what happens
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    if we put a positive test charge
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    in an electric field
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    and that test charge is going to fill a
  • 00:01:09
    force that will accelerate it in the
  • 00:01:12
    same direction as the electric field
  • 00:01:18
    now what about a negative test charge
  • 00:01:21
    what's going to happen if we put it
  • 00:01:23
    in an electric field
  • 00:01:25
    a negative test charge
  • 00:01:28
    will fill a force that will accelerate
  • 00:01:30
    it
  • 00:01:30
    in the opposite direction to the
  • 00:01:32
    electric field
  • 00:01:34
    so make sure you understand that
  • 00:01:36
    positive charges
  • 00:01:38
    they accelerate in the same direction as
  • 00:01:40
    the electric field the force and
  • 00:01:42
    electric field vectors will be in the
  • 00:01:43
    same direction
  • 00:01:44
    but for negative
  • 00:01:46
    charges
  • 00:01:47
    they will fill a force that will
  • 00:01:48
    accelerate it in the opposite direction
  • 00:01:50
    of the electric field
  • 00:01:57
    now as we said before an electric field
  • 00:02:00
    can exert a force
  • 00:02:02
    on
  • 00:02:03
    any type of charge a positive charge or
  • 00:02:05
    a negative charge
  • 00:02:06
    but it turns out that charges
  • 00:02:08
    can also create their own electric
  • 00:02:10
    fields
  • 00:02:14
    the electric field created by a positive
  • 00:02:16
    charge
  • 00:02:17
    extends in all directions
  • 00:02:19
    away from the positive charge
  • 00:02:28
    the electric field created by a negative
  • 00:02:30
    charge
  • 00:02:32
    extends in all directions
  • 00:02:34
    toward the negative charge
  • 00:02:36
    so it's going inward towards the
  • 00:02:38
    negative charge in all directions
  • 00:02:45
    now let's talk about the equation that
  • 00:02:47
    will help us to calculate the electric
  • 00:02:48
    field
  • 00:02:50
    created by a point charge
  • 00:02:54
    so let's say we have
  • 00:02:57
    okay let me draw a better circle
  • 00:02:59
    let's say we have this charge
  • 00:03:02
    which we'll call big q
  • 00:03:05
    and we want to calculate the electric
  • 00:03:07
    field
  • 00:03:09
    at some point
  • 00:03:10
    a
  • 00:03:11
    how can we do that and let's say a is
  • 00:03:15
    some distance r
  • 00:03:19
    well we know that
  • 00:03:21
    the electric field at point a will be
  • 00:03:23
    going in this direction
  • 00:03:26
    all you need to do is draw an arrow
  • 00:03:28
    from the positive charge to the point of
  • 00:03:30
    interest and that will give you the
  • 00:03:31
    direction
  • 00:03:32
    of the electric field
  • 00:03:36
    now let's say
  • 00:03:37
    if we place a tiny positive charge
  • 00:03:44
    at point a so let me just erase point a
  • 00:03:52
    so this tiny positive charge will be
  • 00:03:54
    considered our test charge the reason
  • 00:03:56
    why it has to be tiny because if it's
  • 00:03:59
    large enough
  • 00:04:01
    it will affect
  • 00:04:02
    q
  • 00:04:04
    big q will be repelled by little q but
  • 00:04:06
    if the magnitude of little q is very
  • 00:04:08
    very small
  • 00:04:10
    it won't affect big q as much
  • 00:04:12
    and thus it won't affect the electric
  • 00:04:14
    field that is created by big q as much
  • 00:04:17
    so that this equation will work if we
  • 00:04:19
    choose a very tiny positive test charge
  • 00:04:24
    now according to columbus law whenever
  • 00:04:26
    you have two
  • 00:04:29
    charges next to each other they will
  • 00:04:30
    exert a force in each other
  • 00:04:32
    two like charges
  • 00:04:34
    will fill a force that will repel them
  • 00:04:37
    and we can calculate that force using
  • 00:04:39
    this equation it's equal to k times q1
  • 00:04:42
    which we'll call big q
  • 00:04:44
    times q2 which we'll call little q
  • 00:04:47
    over r squared
  • 00:04:48
    where r
  • 00:04:50
    is the distance
  • 00:04:51
    between the two charges
  • 00:04:55
    so what we're going to do now is we're
  • 00:04:56
    going to take
  • 00:04:58
    this equation
  • 00:04:59
    and substitute it for f in that equation
  • 00:05:04
    but first i'm going to rewrite
  • 00:05:06
    this equation as
  • 00:05:09
    f over 1 times 1 over q
  • 00:05:13
    that's the same as f divided by q
  • 00:05:18
    so now let's replace f
  • 00:05:21
    with what we see here
  • 00:05:32
    so this is going to be f i mean it's no
  • 00:05:35
    longer f but it's going to be k
  • 00:05:38
    big q times little q
  • 00:05:40
    over r squared
  • 00:05:42
    so this whole thing is f
  • 00:05:44
    and then we're going to multiply it by 1
  • 00:05:46
    over q
  • 00:05:49
    and so we could cancel out little q the
  • 00:05:51
    test charge
  • 00:05:53
    thus we get the equation that gives us
  • 00:05:55
    the electric field or the magnitude of
  • 00:05:58
    the electric field
  • 00:06:00
    for
  • 00:06:01
    a point charge capital q so it's k q
  • 00:06:05
    over r squared
  • 00:06:12
    so if we have our charge
  • 00:06:15
    q and we wish to calculate the electric
  • 00:06:17
    field
  • 00:06:18
    at point a
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    we could use this formula
  • 00:06:27
    k
  • 00:06:28
    is 9
  • 00:06:30
    times 10 to the 9
  • 00:06:32
    newtons
  • 00:06:33
    times square meters
  • 00:06:35
    over square coulombs
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    q is the magnitude of the charge in
  • 00:06:40
    columns
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    now remember one micro coulomb
  • 00:06:43
    is one times ten to the minus six
  • 00:06:45
    coulombs
  • 00:06:46
    a nano clue is
  • 00:06:48
    10 to the negative nine clues and a
  • 00:06:50
    milli clue is 10 to the minus three
  • 00:06:52
    columns
  • 00:06:55
    and r is going to be the distance
  • 00:06:57
    in meters
  • 00:07:03
    now let's say we have a positive charge
  • 00:07:05
    here
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    and
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    we have point a
  • 00:07:13
    let's say point b
  • 00:07:18
    point c
  • 00:07:21
    and point d
  • 00:07:23
    determine the electric field
  • 00:07:26
    created by
  • 00:07:27
    the charge q
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    at points a b c and d
  • 00:07:33
    so at point a
  • 00:07:35
    the electric field in order to draw it
  • 00:07:37
    we're going to draw starting from the
  • 00:07:38
    positive charge
  • 00:07:40
    towards point a
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    and so the electric field is going north
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    for point b we need to draw towards
  • 00:07:48
    point b
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    so the electric field the direction
  • 00:07:52
    is west
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    for point d it's going to be at an angle
  • 00:08:00
    so it appears to be going in the
  • 00:08:02
    the northwest direction
  • 00:08:06
    and for point c
  • 00:08:07
    just draw towards point c
  • 00:08:09
    so it's going in uh
  • 00:08:12
    oh wait i have to make a correction that
  • 00:08:14
    is not northwest that is north east
  • 00:08:19
    and this is southeast
  • 00:08:23
    now let's do something similar but with
  • 00:08:24
    a negative charge
  • 00:08:28
    so for the sake of practice
  • 00:08:30
    go ahead and determine
  • 00:08:32
    the direction of the electric field
  • 00:08:34
    at the following points
  • 00:08:36
    so point a
  • 00:08:38
    point b
  • 00:08:41
    point c
  • 00:08:44
    and point d
  • 00:08:47
    so this time the electric field is going
  • 00:08:49
    to point toward the negative charge so
  • 00:08:51
    we're going to draw it from the point to
  • 00:08:53
    the negative charge
  • 00:08:56
    so this electric field vector
  • 00:08:58
    is going
  • 00:09:00
    west
  • 00:09:02
    and then here we're going to draw it
  • 00:09:04
    from v towards the negative charge
  • 00:09:06
    so this is going in the south direction
  • 00:09:10
    and then d to the negative charge
  • 00:09:14
    here it's going east
  • 00:09:18
    and then from c to the negative charge
  • 00:09:20
    this electric field vector is pointed in
  • 00:09:24
    the northwest direction
  • 00:09:28
    so that's how we can determine the
  • 00:09:30
    direction
  • 00:09:31
    of any electric field vector that is
  • 00:09:33
    created by some type of charge either a
  • 00:09:36
    positive or negative charge and we could
  • 00:09:38
    determine the direction of that vector
  • 00:09:40
    at any point
  • 00:09:41
    using examples such as these
  • 00:09:43
    but now let's focus on some word
  • 00:09:44
    problems
  • 00:09:46
    number one
  • 00:09:48
    a force of 100 newtons is directed north
  • 00:09:52
    on a negative 20 micro coulomb point
  • 00:09:54
    charge
  • 00:09:55
    what is the magnitude and direction of
  • 00:09:58
    the electric field at this point
  • 00:10:01
    so feel free to pause the video and work
  • 00:10:03
    on this example problem
  • 00:10:06
    so let's begin by drawing a picture
  • 00:10:09
    so let's draw
  • 00:10:10
    our negative point charge
  • 00:10:13
    now there is an electric force that is
  • 00:10:15
    acting on this charge and that electric
  • 00:10:17
    force is directed north
  • 00:10:23
    what is the direction of the electric
  • 00:10:25
    field
  • 00:10:27
    now if you recall
  • 00:10:29
    if we have an electric field pointing
  • 00:10:31
    east
  • 00:10:32
    and if we were to place a positive
  • 00:10:35
    test charge in that electric field
  • 00:10:37
    it will fill a force
  • 00:10:39
    in the same direction as the electric
  • 00:10:40
    field
  • 00:10:41
    but a negative charge
  • 00:10:43
    will fill a force that will accelerate
  • 00:10:44
    in the opposite direction
  • 00:10:48
    so if the electric field
  • 00:10:50
    is
  • 00:10:53
    due east
  • 00:10:54
    the electric force acting on the
  • 00:10:55
    negative charge
  • 00:10:57
    will be west
  • 00:10:59
    so if the force
  • 00:11:01
    is north
  • 00:11:02
    what is the direction of the electric
  • 00:11:04
    field actin
  • 00:11:05
    on this charge q
  • 00:11:07
    it has to be in the other direction
  • 00:11:09
    it has to be south
  • 00:11:12
    so that is the direction of the electric
  • 00:11:14
    field
  • 00:11:16
    it's direct itself
  • 00:11:19
    now if we want an angle
  • 00:11:21
    we can draw this this is 0 90
  • 00:11:24
    180
  • 00:11:26
    270.
  • 00:11:28
    so the electric field is directed along
  • 00:11:29
    the negative y-axis
  • 00:11:32
    so we could say that
  • 00:11:33
    it's at an angle
  • 00:11:34
    of
  • 00:11:35
    270 degrees
  • 00:11:38
    relative to the positive x-axis
  • 00:11:41
    and that is in a counterclockwise
  • 00:11:42
    direction
  • 00:11:44
    so those are the ways in which we can
  • 00:11:46
    describe
  • 00:11:46
    the direction
  • 00:11:49
    of the electric field we could say it's
  • 00:11:51
    due south
  • 00:11:52
    or it's at an angle of 270 degrees
  • 00:11:56
    so now that we have the direction
  • 00:11:58
    of the electric field let's calculate
  • 00:12:00
    the magnitude
  • 00:12:01
    of the electric field
  • 00:12:04
    so we could use this formula we can take
  • 00:12:06
    the electric force and divide it by the
  • 00:12:08
    magnitude of the charge
  • 00:12:10
    the force acting on it is 100
  • 00:12:13
    newtons
  • 00:12:17
    the magnitude of the charge we don't
  • 00:12:19
    need to worry about the negative sign
  • 00:12:20
    since we already know the direction
  • 00:12:22
    the charge is 20 micro coulombs
  • 00:12:25
    and we can replace
  • 00:12:27
    micro with 10 to the six i mean 10 to
  • 00:12:29
    the minus six
  • 00:12:32
    now let's do some algebra let's see if
  • 00:12:33
    we can get this answer without the use
  • 00:12:36
    of a calculator
  • 00:12:37
    we could divide a hundred by twenty
  • 00:12:40
    how many twenties would make up a
  • 00:12:42
    hundred bucks that's gonna be five
  • 00:12:43
    twenty so a hundred divided by twenty is
  • 00:12:45
    five
  • 00:12:47
    now the ten to the minus six if we move
  • 00:12:49
    it to the top
  • 00:12:51
    the negative exponent will become a
  • 00:12:53
    positive exponent so the negative six
  • 00:12:55
    will change to positive six
  • 00:12:57
    and thus the answer is going to be five
  • 00:13:00
    times ten to the sixth
  • 00:13:01
    and the units in newtons per coulomb
  • 00:13:05
    so that is the electric field that's
  • 00:13:07
    acting on this negative charge
  • 00:13:11
    number two
  • 00:13:12
    a positive charge of 50 micro coulombs
  • 00:13:15
    is placed in an electric field of 50 000
  • 00:13:19
    newtons per column directed upward
  • 00:13:22
    what mass
  • 00:13:23
    should the charge have to remain
  • 00:13:24
    suspended in the air
  • 00:13:28
    well let's talk about how we can create
  • 00:13:30
    such an electric field
  • 00:13:34
    we can create this electric field using
  • 00:13:36
    a battery
  • 00:13:38
    and two
  • 00:13:39
    parallel metal plates
  • 00:13:46
    so this is the electrical symbol of a
  • 00:13:48
    battery
  • 00:13:50
    we're going to connect it to these two
  • 00:13:51
    plates
  • 00:13:55
    this is the negative side of the battery
  • 00:13:58
    and this is the positive side
  • 00:14:00
    so this plate is going to acquire a
  • 00:14:02
    positive charge
  • 00:14:05
    and this plate
  • 00:14:08
    is going to acquire a negative charge
  • 00:14:13
    so if we choose a high enough voltage
  • 00:14:16
    and if the distance between the two
  • 00:14:18
    plates is just right
  • 00:14:20
    we can adjust it such that we get an
  • 00:14:22
    electric field
  • 00:14:23
    of 50 000 newtons per coulomb
  • 00:14:26
    now we want the electric field to be
  • 00:14:28
    directed upward
  • 00:14:30
    so we want the positively charged plate
  • 00:14:32
    to be at the bottom the negatively
  • 00:14:33
    charged plate to be at the top remember
  • 00:14:36
    the electric field always extends away
  • 00:14:38
    from a positive charge and points
  • 00:14:40
    towards a negative charge
  • 00:14:42
    so this is how we can create a uniform
  • 00:14:44
    electric field using two parallel plate
  • 00:14:47
    i mean using two
  • 00:14:49
    parallel metal plates
  • 00:14:51
    now let's place our positive charge
  • 00:14:54
    in the middle between these two plates
  • 00:14:58
    if this charge has the right mass
  • 00:15:01
    it can actually remain suspended in the
  • 00:15:03
    air
  • 00:15:04
    now let's talk about it
  • 00:15:06
    so we have an electric field that is
  • 00:15:07
    directed upward
  • 00:15:10
    and it's acting on its positive charge
  • 00:15:13
    what will be the direction of the
  • 00:15:15
    electric force
  • 00:15:16
    on its charge
  • 00:15:18
    the electric force will be the same
  • 00:15:20
    direction as the electric field for a
  • 00:15:22
    positive charge
  • 00:15:24
    so the electric force will be directed
  • 00:15:26
    upward
  • 00:15:27
    now gravity
  • 00:15:28
    gravity likes to bring things down
  • 00:15:31
    so gravity is going to exert a weight
  • 00:15:33
    force
  • 00:15:34
    on this positive charge
  • 00:15:36
    bringing it down in the negative y
  • 00:15:38
    direction
  • 00:15:39
    in order for this charge to remain
  • 00:15:41
    suspended in the air
  • 00:15:43
    the electric force
  • 00:15:45
    that wants to accelerate the charge
  • 00:15:47
    towards the negatively charged pay
  • 00:15:50
    that has to be exactly equal to the
  • 00:15:52
    gravitational force or the weight force
  • 00:15:54
    that wants to
  • 00:15:56
    bring the positive charge down towards
  • 00:15:58
    the
  • 00:15:58
    towards this plate
  • 00:16:00
    so if we can get these two forces to
  • 00:16:02
    equal each other
  • 00:16:03
    then the positive charge
  • 00:16:05
    will remain suspended in the air
  • 00:16:10
    so let's write
  • 00:16:12
    the forces in the y direction the sum of
  • 00:16:15
    the forces in the y direction
  • 00:16:17
    is equal to the electric force this is
  • 00:16:19
    positive
  • 00:16:20
    and the weight force is negative because
  • 00:16:21
    it's going in a negative y direction
  • 00:16:25
    and we want the the sum of the forces in
  • 00:16:27
    the y direction to be zero so that
  • 00:16:29
    there's no net acceleration so it
  • 00:16:31
    remains suspended in the air
  • 00:16:33
    moving this term to the other side
  • 00:16:35
    we see that the electric force has to
  • 00:16:38
    equal the weight force
  • 00:16:42
    now
  • 00:16:43
    the electric field is equal to f over q
  • 00:16:47
    multiplying both sides by q
  • 00:16:49
    we can see that f
  • 00:16:51
    the electric force is
  • 00:16:53
    the electric field times q
  • 00:16:55
    so let's replace the electric force with
  • 00:16:57
    e times q
  • 00:17:00
    now the wave force is simply mg
  • 00:17:05
    so now we have everything that we need
  • 00:17:07
    in order to calculate the mass of this
  • 00:17:08
    charge
  • 00:17:09
    let's divide both sides by g
  • 00:17:13
    so the mass of the charge is going to be
  • 00:17:15
    equal to the electric field times the
  • 00:17:17
    magnitude of the charge
  • 00:17:19
    divided by the gravitational
  • 00:17:20
    acceleration
  • 00:17:23
    the electric field is 50 000
  • 00:17:27
    newtons per coulomb
  • 00:17:29
    the magnitude of the charge we're
  • 00:17:30
    dealing with uh a 50 microclimate charge
  • 00:17:34
    so it's going to be 50
  • 00:17:36
    times 10 to the negative 6 coulombs
  • 00:17:39
    and then we're going to divide that by
  • 00:17:41
    the gravitational acceleration of 9.8
  • 00:17:44
    meters per second squared
  • 00:17:47
    so it's 50 000 times 50 times 10 to the
  • 00:17:50
    minus 6 divided by 9.8
  • 00:17:53
    and you should get
  • 00:17:54
    .255
  • 00:17:57
    kilograms
  • 00:18:01
    so an electric field of 50 000 newtons
  • 00:18:04
    per coulomb
  • 00:18:05
    can suspend
  • 00:18:07
    a positive charge with a mass of 0.255
  • 00:18:10
    kilograms or 255 grams
  • 00:18:14
    it can suspend it in the air
  • 00:18:17
    if the mass is greater than this number
  • 00:18:19
    then the charge will fall down
  • 00:18:22
    if it's too light if the mass is less
  • 00:18:23
    than this number
  • 00:18:24
    then the charge will accelerate towards
  • 00:18:27
    the negatively charged plate
  • 00:18:29
    so
  • 00:18:30
    the mass has to be at the right number
  • 00:18:32
    in order for it to remain suspended
  • 00:18:35
    number three
  • 00:18:36
    an electron is released from rust in a
  • 00:18:39
    uniform electric field and accelerates
  • 00:18:42
    to the east
  • 00:18:43
    at a rate of four times ten to the sixth
  • 00:18:45
    meters per second squared
  • 00:18:47
    what is the magnitude and direction of
  • 00:18:50
    the electric field
  • 00:18:52
    so let's draw a picture
  • 00:18:53
    so first let's draw
  • 00:18:55
    our electron
  • 00:19:00
    and then it is
  • 00:19:01
    accelerating towards the east
  • 00:19:07
    now according to newton's second law the
  • 00:19:10
    net force is equal to the mass times the
  • 00:19:12
    acceleration
  • 00:19:13
    the net force is in the same direction
  • 00:19:16
    as the acceleration
  • 00:19:18
    so the electric force exerted by this
  • 00:19:20
    electron due to the electric field is
  • 00:19:22
    going to be due east as well
  • 00:19:26
    now if the electric force is east
  • 00:19:28
    what is the direction of the electric
  • 00:19:30
    field
  • 00:19:31
    it's going to have to be west
  • 00:19:34
    let me put it here
  • 00:19:37
    when dealing with a negative charge the
  • 00:19:38
    direction of the force and electric
  • 00:19:40
    field they will be opposite to each
  • 00:19:42
    other
  • 00:19:44
    so now that we have the direction of the
  • 00:19:45
    electric field
  • 00:19:47
    let's focus on getting the magnitude of
  • 00:19:49
    the electric field
  • 00:19:53
    so from the last problem we saw that the
  • 00:19:54
    electric force
  • 00:19:56
    is equal to the electric field times the
  • 00:19:58
    charge
  • 00:19:59
    and
  • 00:20:00
    using newton's second law we can replace
  • 00:20:03
    the force with
  • 00:20:04
    the mass times the acceleration
  • 00:20:08
    so now we can calculate the electric
  • 00:20:09
    field if we divide both sides by q
  • 00:20:14
    so for this problem
  • 00:20:16
    the magnitude of the electric field is
  • 00:20:18
    going to be the mass times the
  • 00:20:19
    acceleration which is the force
  • 00:20:21
    divided by the charge
  • 00:20:25
    so the mass is
  • 00:20:27
    what is the mass of an electron if you
  • 00:20:29
    look it up
  • 00:20:30
    the mass of an electron is 9.11
  • 00:20:33
    times 10
  • 00:20:34
    to the negative 31 kilograms
  • 00:20:38
    the acceleration given to us in this
  • 00:20:40
    problem
  • 00:20:41
    is 4
  • 00:20:43
    times 10 to the 6 meters per second
  • 00:20:45
    squared
  • 00:20:48
    and then we're going to divide that by
  • 00:20:51
    the magnitude of the charge
  • 00:20:53
    so what is the charge of an electron
  • 00:20:56
    the charge of an electron is negative
  • 00:20:59
    1.602
  • 00:21:00
    times 10 to the negative 19 coulombs
  • 00:21:04
    so these are some numbers that you want
  • 00:21:05
    to be familiar with
  • 00:21:09
    so let's put that here
  • 00:21:18
    so let's go ahead and plug in these
  • 00:21:20
    numbers
  • 00:21:27
    by the way don't worry about the
  • 00:21:28
    negative sign for q
  • 00:21:30
    it's not going to be relevant here
  • 00:21:33
    so the magnitude of the electric field
  • 00:21:34
    is going to be 2.27
  • 00:21:38
    times 10 to the negative 5
  • 00:21:41
    newtons per coulomb
  • 00:21:44
    so this is the answer for this problem
  • 00:21:47
    and the direction
  • 00:21:48
    is
  • 00:21:49
    west
  • 00:21:53
    so make sure you write down these
  • 00:21:54
    numbers
  • 00:21:56
    the mass of an electron as we've
  • 00:21:58
    considered is 9.11
  • 00:22:01
    times 10 to the negative 31 kilograms
  • 00:22:04
    the mass of a proton
  • 00:22:06
    is 1.67
  • 00:22:08
    times 10
  • 00:22:09
    to the negative 27
  • 00:22:11
    kilograms
  • 00:22:13
    the charge of an electron
  • 00:22:15
    it's going to be a negative
  • 00:22:18
    1.602
  • 00:22:20
    times 10 to the negative 19 coulombs
  • 00:22:24
    the charge of the proton
  • 00:22:26
    it's going to have the same magnitude
  • 00:22:27
    but the opposite charge is going to be
  • 00:22:29
    positive
  • 00:22:30
    1.602
  • 00:22:32
    times 10 to the negative 19 coulombs
  • 00:22:35
    so those are some numbers that you want
  • 00:22:36
    to be familiar with
  • 00:22:38
    when working on problems associated with
  • 00:22:40
    electric fields if you're dealing with
  • 00:22:42
    protons and electrons
  • 00:22:44
    number four
  • 00:22:45
    a 40 micro coulomb point charge is
  • 00:22:47
    placed at the origin
  • 00:22:49
    calculate the magnitude and direction of
  • 00:22:51
    the electric field created by the point
  • 00:22:53
    charge
  • 00:22:54
    at the following locations
  • 00:22:58
    so let's draw the point charge first
  • 00:23:07
    now point p
  • 00:23:08
    is five meters away
  • 00:23:11
    from the point charge along the x-axis
  • 00:23:15
    so this is point p
  • 00:23:17
    and this is five meters away
  • 00:23:21
    what is the electric field at point p
  • 00:23:23
    well we know the direction
  • 00:23:25
    the direction of the electric field
  • 00:23:27
    is going to be
  • 00:23:28
    east
  • 00:23:31
    if we draw it from the positive charge
  • 00:23:32
    towards point p
  • 00:23:35
    now to calculate the magnitude of the
  • 00:23:38
    electric field it's going to be k
  • 00:23:40
    q
  • 00:23:41
    divided by r squared
  • 00:23:43
    so k is 9 times 10 to 9 and then it's
  • 00:23:47
    newtons
  • 00:23:48
    times square meters
  • 00:23:50
    over square columns
  • 00:23:53
    and i'm running out of space so i'm not
  • 00:23:55
    going to put the units here i'm just
  • 00:23:56
    going to write 9 times 10 to the nine
  • 00:23:58
    q
  • 00:23:59
    is 40 micro clones so 40 times 10 to the
  • 00:24:02
    minus 6 coulombs
  • 00:24:04
    r is in meters
  • 00:24:06
    r is the distance between a charge and
  • 00:24:08
    the point of interest
  • 00:24:09
    so that's five meters
  • 00:24:16
    so it's going to be nine times 10 to the
  • 00:24:17
    9
  • 00:24:18
    times 40 times 10 to the minus 6
  • 00:24:21
    over 5 squared
  • 00:24:23
    and so the electric field is going to be
  • 00:24:25
    14
  • 00:24:27
    400
  • 00:24:29
    newtons per coulomb
  • 00:24:31
    so that's the magnitude of the electric
  • 00:24:32
    field at point p
  • 00:24:34
    and this is the direction
  • 00:24:36
    for those of you who want to understand
  • 00:24:37
    how the units work
  • 00:24:39
    here it is so k is in newtons
  • 00:24:42
    times square meters over square coulombs
  • 00:24:46
    q is in coulombs and r is in meters so
  • 00:24:48
    we have square meters
  • 00:24:50
    as you can see square meters cancel
  • 00:24:53
    now coulomb squared we can write that as
  • 00:24:55
    coulomb times coulomb
  • 00:24:58
    so we can cancel
  • 00:25:00
    one of the coulomb units
  • 00:25:02
    thus we're left with newtons per coulomb
  • 00:25:05
    which is what we have here
  • 00:25:09
    so that's it for part a so it's 14 400
  • 00:25:12
    newtons per coulomb directed east
  • 00:25:22
    now let's move on to part b
  • 00:25:24
    so let's calculate the electric field
  • 00:25:26
    at point s
  • 00:25:29
    so
  • 00:25:30
    q is at the origin
  • 00:25:32
    point s is three meters
  • 00:25:35
    east of q
  • 00:25:37
    and then four meters north from that
  • 00:25:39
    point
  • 00:25:47
    so s
  • 00:25:49
    is at this position here
  • 00:25:52
    the electric field
  • 00:25:54
    can be drawn from q to s
  • 00:25:57
    so the electric field is going in that
  • 00:25:58
    direction it's going in the
  • 00:26:01
    the northeast direction
  • 00:26:04
    let's calculate the magnitude of the
  • 00:26:05
    electric field
  • 00:26:07
    so let's use this formula again it's
  • 00:26:09
    going to be k
  • 00:26:10
    q
  • 00:26:11
    over r squared
  • 00:26:14
    so k doesn't change it's a constant it's
  • 00:26:16
    9 times 10 to 9.
  • 00:26:18
    q is still the same
  • 00:26:20
    it's 40
  • 00:26:21
    times 10 to the minus
  • 00:26:23
    but r is different
  • 00:26:26
    r is no longer
  • 00:26:30
    the value that we have here
  • 00:26:32
    but in actuality it turns out r is the
  • 00:26:34
    same
  • 00:26:36
    r
  • 00:26:37
    is the distance between the charge and
  • 00:26:39
    the point so we need to use
  • 00:26:41
    the pythagorean theorem to calculate the
  • 00:26:43
    hypotenuse of that right triangle
  • 00:26:45
    so this is a b and this is c
  • 00:26:48
    so c squared is equal to a squared plus
  • 00:26:50
    b squared
  • 00:26:52
    a is three
  • 00:26:54
    b
  • 00:26:55
    is four
  • 00:26:57
    three squared is nine four squared is
  • 00:26:59
    sixteen
  • 00:27:00
    and then nine plus sixteen is twenty
  • 00:27:02
    five
  • 00:27:03
    taking the square root of both sides
  • 00:27:06
    we get 5 again
  • 00:27:10
    so by coincidence the electric field is
  • 00:27:13
    going to have the same magnitude
  • 00:27:15
    as it did in part a
  • 00:27:19
    which was uh 14
  • 00:27:22
    400
  • 00:27:23
    newtons per coulomb
  • 00:27:27
    what's going to be different though is
  • 00:27:28
    the direction of the electric field
  • 00:27:30
    vector
  • 00:27:32
    how can we determine the direction
  • 00:27:34
    we know it's somewhat in the northeast
  • 00:27:36
    direction but not necessarily at a 45
  • 00:27:38
    degree angle
  • 00:27:39
    what we need to do is calculate theta
  • 00:27:43
    so
  • 00:27:44
    perhaps you remember from trigonometry
  • 00:27:45
    sokotova
  • 00:27:48
    if we focus on
  • 00:27:49
    the torah part that tells us tangent
  • 00:27:52
    tangent of theta is equal to the
  • 00:27:54
    opposite side
  • 00:27:56
    divided by the adjacent side
  • 00:28:01
    so tangent of the angle theta is going
  • 00:28:03
    to be equal to four over three
  • 00:28:05
    to calculate theta we need to take the
  • 00:28:07
    arc tangent
  • 00:28:08
    or the inverse tangent of four over
  • 00:28:10
    three
  • 00:28:12
    go ahead and type in your calculator and
  • 00:28:13
    make sure it's in degree mode so arc
  • 00:28:16
    tangent 4 3
  • 00:28:18
    is
  • 00:28:19
    53.1 degrees
  • 00:28:22
    so that is the angle
  • 00:28:24
    with respect to
  • 00:28:25
    the positive x-axis
  • 00:28:28
    it's 53.1
  • 00:28:30
    so we could say that this is the
  • 00:28:32
    electric field vector
  • 00:28:33
    at an angle of 53.1 degrees with respect
  • 00:28:37
    to the x-axis or we could say
  • 00:28:39
    53.1 degrees
  • 00:28:41
    north of east
  • 00:28:44
    because here we're starting from east
  • 00:28:47
    and we're heading towards the north
  • 00:28:49
    direction so it's 53.1
  • 00:28:52
    north of east
  • 00:28:54
    so that's it for part b so that's how we
  • 00:28:57
    can calculate the magnitude of the
  • 00:28:59
    electric field vector
  • 00:29:01
    and also its direction using arctangent
  • 00:29:05
    number five
  • 00:29:06
    an electron initially at rest
  • 00:29:09
    is placed in an electric field of two
  • 00:29:11
    times ten to the four
  • 00:29:13
    newtons per coulomb directed to the west
  • 00:29:17
    the distance between the plates is one
  • 00:29:19
    centimeter
  • 00:29:20
    what is the acceleration of the electron
  • 00:29:22
    due to the electric field
  • 00:29:25
    so the electric field will emanate away
  • 00:29:27
    from the positive charge and will point
  • 00:29:29
    towards the negative charge
  • 00:29:31
    so as we can see the electric field
  • 00:29:34
    is directed west
  • 00:29:39
    now what effect will it have on the
  • 00:29:40
    electron
  • 00:29:42
    a negatively charged particle will fill
  • 00:29:44
    a force that will accelerate it in the
  • 00:29:46
    opposite direction to the electric field
  • 00:29:50
    so the electron is going to shoot out of
  • 00:29:52
    this uh between these two parallel
  • 00:29:54
    plates
  • 00:29:55
    let's calculate the acceleration
  • 00:29:59
    we know the force acting on a charged
  • 00:30:01
    particle is equal to the electric field
  • 00:30:03
    times the charge of that particle
  • 00:30:05
    and since this is the only force acting
  • 00:30:07
    on the electron in the x direction
  • 00:30:10
    then that force
  • 00:30:11
    is going to be
  • 00:30:13
    the electric force so we can replace the
  • 00:30:14
    net force with m a based on newton's
  • 00:30:16
    second law
  • 00:30:17
    so m a is equal to e times q
  • 00:30:21
    and to solve for a we're going to divide
  • 00:30:22
    both sides by m
  • 00:30:24
    so for part a the acceleration
  • 00:30:27
    is going to be the force which is e
  • 00:30:28
    times q
  • 00:30:31
    divided by the mass
  • 00:30:34
    so we have an electric field of 2 times
  • 00:30:37
    10 to the 4
  • 00:30:39
    newtons per coulomb
  • 00:30:42
    and the charge of an electron is 1.602
  • 00:30:46
    times 10 to the negative 19 coulombs
  • 00:30:49
    we're not going to worry about the
  • 00:30:50
    negative sign the mass of an electron is
  • 00:30:53
    9.11
  • 00:30:56
    times 10 to the negative 31 kilograms
  • 00:30:59
    so let's go ahead and plug this in
  • 00:31:08
    so the acceleration
  • 00:31:10
    is 3.517
  • 00:31:14
    times 10 to the 15
  • 00:31:16
    meters per second squared
  • 00:31:19
    so that's going to be the acceleration
  • 00:31:23
    when the electron
  • 00:31:24
    leaves
  • 00:31:26
    the negatively charged plate
  • 00:31:29
    so now let's move on to part b
  • 00:31:31
    what would be the speed of the electron
  • 00:31:33
    after it leaves the hole
  • 00:31:36
    so how can we get that
  • 00:31:37
    so now we need to go back to kinematics
  • 00:31:41
    the electron is initially at rest so v
  • 00:31:43
    initial is zero
  • 00:31:44
    we're looking for the final speed
  • 00:31:46
    so we'll put a question mark
  • 00:31:48
    and we know the distance between the
  • 00:31:49
    plates
  • 00:31:50
    it's approximately
  • 00:31:52
    it's one centimeter
  • 00:31:54
    so what kinematic formula has
  • 00:31:56
    acceleration v initial v final and d
  • 00:32:00
    it's going to be this one
  • 00:32:02
    v final squared is equal to v initial
  • 00:32:04
    squared
  • 00:32:05
    plus 2 a d
  • 00:32:07
    so to solve for the final speed we
  • 00:32:09
    simply need to take the square root of
  • 00:32:11
    both sides
  • 00:32:12
    so v final is equal to this
  • 00:32:17
    the initial speed is zero
  • 00:32:19
    this is going to be two
  • 00:32:21
    times the acceleration which is
  • 00:32:24
    3.517
  • 00:32:26
    times 10 to the 15.
  • 00:32:30
    and the distance between its plates is
  • 00:32:32
    one centimeter so
  • 00:32:34
    if we convert one centimeter into meters
  • 00:32:36
    we need to divide by 100 and it's going
  • 00:32:38
    to be 0.01 meters
  • 00:32:50
    and so we're going to get the square
  • 00:32:51
    root of 7.034
  • 00:32:55
    times ten to the thirteen
  • 00:33:00
    and so the final speed
  • 00:33:02
    is about
  • 00:33:04
    eight million and three hundred eighty
  • 00:33:06
    six thousand eight ninety four point
  • 00:33:07
    five
  • 00:33:08
    so we can round that to
  • 00:33:10
    let's say 8.39
  • 00:33:14
    and this is times 10 to the sixth
  • 00:33:17
    meters per second
  • 00:33:19
    so that's how fast the electron is going
  • 00:33:20
    to be moving when it leaves the hole
  • 00:33:24
    number six
  • 00:33:25
    a 200 micro coulomb charge is placed at
  • 00:33:28
    the origin
  • 00:33:29
    and a negative 300 micro coulomb charge
  • 00:33:32
    is placed one meter to the right of it
  • 00:33:36
    what is the magnitude and direction
  • 00:33:38
    of the electric field midway between the
  • 00:33:40
    two charges and then for the second part
  • 00:33:42
    30 centimeters to the right of the
  • 00:33:44
    negative charge
  • 00:33:47
    so let's start with the first part of
  • 00:33:48
    the problem
  • 00:33:54
    let's begin by drawing a picture
  • 00:33:59
    so this is going to be the first charge
  • 00:34:01
    we'll call it q1
  • 00:34:02
    and the second charge q2
  • 00:34:05
    so these two charges are separated by a
  • 00:34:07
    distance
  • 00:34:08
    of one meter
  • 00:34:09
    and we want to determine
  • 00:34:11
    the magnitude and the direction of the
  • 00:34:13
    electric field midway between the two
  • 00:34:15
    charges
  • 00:34:17
    so that's going to be at this point
  • 00:34:22
    how can we do that
  • 00:34:24
    well we need to determine the direction
  • 00:34:26
    of each electric field at that point
  • 00:34:29
    q1
  • 00:34:30
    is going to create an electric field
  • 00:34:32
    called e1 which will be directed east
  • 00:34:35
    now q2 will create an electric field e2
  • 00:34:39
    which starts from the point but points
  • 00:34:41
    towards the negative charge
  • 00:34:43
    and that's going to be directed east as
  • 00:34:44
    well
  • 00:34:46
    so remember the electric field created
  • 00:34:49
    by a positive charge extends away from
  • 00:34:51
    the positive charge
  • 00:34:53
    but the electric field created by a
  • 00:34:54
    negative charge points towards the
  • 00:34:56
    negative charge
  • 00:34:59
    so at the center e1 and e2 they are in
  • 00:35:02
    the same direction
  • 00:35:04
    so the net electric field
  • 00:35:06
    is going to be e1 plus e2
  • 00:35:10
    along the x-axis or the horizontal axis
  • 00:35:13
    both of these are positive because
  • 00:35:15
    they're going in the positive x
  • 00:35:16
    direction
  • 00:35:24
    e1 is k
  • 00:35:26
    times q1
  • 00:35:27
    over r1 squared e2 is k times q2
  • 00:35:34
    over r2 squared
  • 00:35:38
    now what's r1 and r2
  • 00:35:40
    so r1 is the distance between q1 and the
  • 00:35:43
    point of interest
  • 00:35:44
    r2 is the distance between q2
  • 00:35:47
    and
  • 00:35:48
    the point of interest
  • 00:35:51
    so r1 and r2 they're both half of point
  • 00:35:54
    of one meter which means that r1 and r2
  • 00:35:57
    they're both equal to 0.5
  • 00:36:01
    so because r1 and r2 are the same we can
  • 00:36:04
    simply call it r
  • 00:36:07
    so let's replace r1 with r
  • 00:36:10
    and let's replace r2
  • 00:36:12
    with r as well
  • 00:36:15
    so now we could simplify this equation
  • 00:36:17
    by
  • 00:36:18
    factoring out the gcf the greatest
  • 00:36:20
    common factor which is going to be k
  • 00:36:22
    over r squared
  • 00:36:24
    and then we're left with q1 plus q2
  • 00:36:29
    so this is the formula that we could use
  • 00:36:31
    to calculate the net electric field for
  • 00:36:33
    this particular part in the problem
  • 00:36:35
    now let's go ahead and plug in the
  • 00:36:37
    numbers
  • 00:36:43
    so it's going to be k which is 9 times
  • 00:36:45
    10 to the 9
  • 00:36:47
    over r squared
  • 00:36:49
    r is 0.5
  • 00:36:52
    and then times q1 q1 is 200
  • 00:36:56
    times 10 to the negative six
  • 00:36:58
    now for q2
  • 00:36:59
    we're going to use a positive value not
  • 00:37:02
    negative 300 times 10 to the negative
  • 00:37:04
    six
  • 00:37:05
    because we already know the direction of
  • 00:37:07
    e2 it's going to the right and it's
  • 00:37:09
    going to have a positive value because
  • 00:37:10
    it's heading
  • 00:37:11
    in the eastward direction
  • 00:37:15
    so whenever you're calculating the
  • 00:37:16
    magnitude for electric field or electric
  • 00:37:19
    force
  • 00:37:20
    you don't need to include the negative
  • 00:37:22
    charge
  • 00:37:23
    you could just find the direction based
  • 00:37:24
    on where the arrow is going
  • 00:37:28
    so let's replace q2 with 300
  • 00:37:30
    times 10
  • 00:37:32
    to the negative six coulombs
  • 00:37:38
    so 200 plus 300 this becomes 500 so it's
  • 00:37:43
    9 times 10 to the 9
  • 00:37:45
    times 500
  • 00:37:46
    times 10 to negative 6
  • 00:37:48
    divided by 0.5 squared
  • 00:37:52
    thus the net electric field
  • 00:37:55
    is 18 million which is 1.8
  • 00:37:59
    times 10 to the 7
  • 00:38:01
    newtons per coulomb
  • 00:38:03
    so that's the answer for part a
  • 00:38:08
    now let's move on to part b
  • 00:38:10
    so let's redraw the picture for that
  • 00:38:17
    so here we have our positive charge
  • 00:38:20
    and the negative charge
  • 00:38:24
    so we got q1
  • 00:38:26
    q2 and they're separated by distance of
  • 00:38:28
    one meter
  • 00:38:30
    but 30 centimetres to the right or 0.3
  • 00:38:33
    meters
  • 00:38:34
    we're going to have our new point of
  • 00:38:35
    interest
  • 00:38:36
    and let's call this
  • 00:38:38
    point b
  • 00:38:40
    so we want to determine the net electric
  • 00:38:42
    field at point b
  • 00:38:45
    so e1 the electric field created by q1
  • 00:38:48
    if we draw it from q1 to point b we
  • 00:38:50
    could see that it's going east
  • 00:38:57
    now if we draw an electric field from
  • 00:39:00
    point b to q two
  • 00:39:02
    because it's a negative charge it needs
  • 00:39:04
    to go towards a negative charge it's
  • 00:39:06
    going to the left
  • 00:39:09
    now which of these two electric fields
  • 00:39:12
    is greater e1 or e2 what would you say
  • 00:39:16
    notice that point b is closer
  • 00:39:19
    to the negative charge
  • 00:39:21
    than it is to the positive charge
  • 00:39:23
    so e2 is going to have a greater effect
  • 00:39:26
    than uh i mean q2 is going to have a
  • 00:39:27
    greater effect on point b than q1
  • 00:39:30
    because it's closer
  • 00:39:33
    so if you go back to the equation for
  • 00:39:35
    electric field
  • 00:39:37
    there's two things that the electric
  • 00:39:39
    field depends on
  • 00:39:40
    the magnitude of the charge
  • 00:39:42
    and the inverse square of the distance
  • 00:39:44
    but the distance is square
  • 00:39:46
    so
  • 00:39:47
    the distance has a greater impact than
  • 00:39:49
    the charge but also
  • 00:39:52
    q2 has a bigger charge in q1
  • 00:39:55
    so those are two factors that favor q2
  • 00:39:58
    over q1 point b is closer to q2
  • 00:40:01
    and q2 has a greater
  • 00:40:03
    uh charge magnitude than q1
  • 00:40:07
    so therefore we can conclude that e2 is
  • 00:40:09
    going to be bigger than e1
  • 00:40:14
    now the net electric field
  • 00:40:16
    is going to be positive e1 because it's
  • 00:40:19
    going along the positive x-axis
  • 00:40:21
    and then plus negative e2 because
  • 00:40:24
    that's moving towards the west or the
  • 00:40:26
    negative x-axis
  • 00:40:29
    now if e1 is bigger than e2
  • 00:40:31
    the net electric field will be positive
  • 00:40:34
    if e2 is greater than e1 it's going to
  • 00:40:36
    be negative and we've confirmed that e2
  • 00:40:40
    is going to be bigger so we should get a
  • 00:40:41
    negative value if we get a positive
  • 00:40:43
    value for the net electric field
  • 00:40:45
    we did something wrong
  • 00:40:48
    so let's go ahead and do the math
  • 00:40:54
    so then that electric field
  • 00:40:56
    is going to be k
  • 00:40:59
    q 1
  • 00:41:00
    over r 1 squared
  • 00:41:03
    plus k
  • 00:41:04
    q q2
  • 00:41:06
    over r2 squared
  • 00:41:08
    so this time r1 and r2 will be different
  • 00:41:11
    so we can't simplify
  • 00:41:13
    this process by factoring
  • 00:41:16
    so let's plug in the numbers this is k
  • 00:41:19
    q1 is 200
  • 00:41:21
    oh this should be a negative sign by the
  • 00:41:22
    way
  • 00:41:26
    based on what we have here
  • 00:41:28
    so q1 is 200 times 10 to negative 6.
  • 00:41:32
    now r1
  • 00:41:33
    r1 is the distance between q1
  • 00:41:36
    and point b
  • 00:41:38
    so r1 is going to be the sum of one and
  • 00:41:40
    point three
  • 00:41:42
    thus r1 is 1.3 meters squared
  • 00:41:46
    and then minus
  • 00:41:48
    now because we've considered the
  • 00:41:50
    direction of e2 we've assigned a
  • 00:41:52
    negative value
  • 00:41:53
    we don't need to plug in this negative
  • 00:41:55
    value for q2
  • 00:41:57
    we've already taken that into an account
  • 00:42:00
    so it's to be minus k
  • 00:42:04
    and then q2 we're going to use the
  • 00:42:05
    positive value of q2
  • 00:42:07
    300 times 10 to the minus 6
  • 00:42:11
    and r2 that's the distance between q2
  • 00:42:14
    and point b
  • 00:42:15
    that's 0.3 meters
  • 00:42:24
    so let's calculate e1 first
  • 00:42:27
    let's focus on this fraction
  • 00:42:30
    nine times ten to the nine times two
  • 00:42:32
    hundred times ten to the negative six
  • 00:42:35
    divided by one point three squared
  • 00:42:38
    that's one point
  • 00:42:40
    zero six five
  • 00:42:41
    times ten to the seven
  • 00:42:43
    newton
  • 00:42:44
    newtons per coulomb
  • 00:42:46
    now focusing on e2
  • 00:42:49
    that's gonna be nine times ten to nine
  • 00:42:50
    times three hundred times ten to
  • 00:42:52
    negative six
  • 00:42:53
    divided by point three squared
  • 00:42:57
    so this is
  • 00:43:00
    3
  • 00:43:01
    times 10 to the 7 newtons per coulomb so
  • 00:43:04
    we can see that this number is bigger
  • 00:43:06
    than that number
  • 00:43:07
    so the net electric field is going to be
  • 00:43:09
    negative
  • 00:43:11
    1.065
  • 00:43:13
    minus 3.
  • 00:43:18
    wait something is wrong
  • 00:43:21
    let me double check my work
  • 00:43:34
    this should be times ten to the six not
  • 00:43:37
    ten to the seven
  • 00:43:42
    that's one million sixty five thousand
  • 00:43:44
    and eighty eight so that's one point
  • 00:43:46
    zero six five times ten to six
  • 00:43:54
    e2 is 30 million which is three times
  • 00:43:56
    ten to seven
  • 00:43:58
    so now if we subtract those two numbers
  • 00:44:02
    we get this answer
  • 00:44:04
    negative
  • 00:44:05
    two point
  • 00:44:07
    eight nine
  • 00:44:08
    times
  • 00:44:10
    so it's 28 million nine hundred thirty
  • 00:44:12
    five thousand
  • 00:44:13
    so it's negative two point eight nine
  • 00:44:15
    times ten to the seven
  • 00:44:17
    newtons per coulomb
  • 00:44:19
    so we can see why it's negative e2
  • 00:44:22
    is significantly
  • 00:44:24
    larger than e1
  • 00:44:26
    and so that's it for this problem
  • 00:44:28
    number seven
  • 00:44:30
    the electric field at point x
  • 00:44:32
    two meters to the right of a certain
  • 00:44:33
    positive charge is 100 newtons per
  • 00:44:36
    coulomb
  • 00:44:37
    what will be the magnitude of the new
  • 00:44:38
    electric field if the magnitude of the
  • 00:44:40
    positive charge doubles in value
  • 00:44:43
    so let's draw a picture first
  • 00:44:45
    so here is our positive charge
  • 00:44:48
    and let's say this is
  • 00:44:50
    point x
  • 00:44:53
    and the distance between these two
  • 00:44:56
    is two meters
  • 00:45:00
    now at that point the electric field is
  • 00:45:01
    a hundred newtons per coulomb
  • 00:45:04
    when the magnitude of the charge is
  • 00:45:06
    we'll call it q
  • 00:45:09
    but what happens if we double
  • 00:45:11
    the magnitude of the charge
  • 00:45:13
    so the electric field is k
  • 00:45:16
    q over r squared
  • 00:45:18
    if you double q
  • 00:45:21
    the electric field is going to double a
  • 00:45:23
    quick and simple way to get the answer
  • 00:45:25
    is to
  • 00:45:26
    plug in 1 for everything that that
  • 00:45:28
    doesn't change
  • 00:45:29
    q doubles so we're going to plug in 2
  • 00:45:31
    r remains the same
  • 00:45:33
    so the electric field is going to double
  • 00:45:34
    it's going to go from 100 to 200.
  • 00:45:39
    now what about part b
  • 00:45:41
    let's say the distance between the
  • 00:45:43
    charge and point x doubles
  • 00:45:51
    so let's say point x is now over here
  • 00:45:56
    and the magnitude is q
  • 00:45:58
    what will be the new electric field
  • 00:46:04
    so this time q doesn't change
  • 00:46:07
    k is the same so we're going to replace
  • 00:46:08
    it with a 1. everything that doesn't
  • 00:46:10
    change replace it with 1.
  • 00:46:13
    now the distance doubles
  • 00:46:17
    two squared is four so the electric
  • 00:46:19
    field is going to be one-fourth
  • 00:46:21
    of its original value
  • 00:46:24
    one-fourth of a hundred or hundred
  • 00:46:26
    divided by four
  • 00:46:27
    that's 25
  • 00:46:29
    so what you need to take from this is
  • 00:46:31
    that the electric field is weaker
  • 00:46:34
    at a greater distance away from the
  • 00:46:36
    charge
  • 00:46:37
    the closer you move towards the point
  • 00:46:39
    charge
  • 00:46:40
    the greater the electric field will be
  • 00:46:43
    so as the distance increases
  • 00:46:46
    the electric field decreases
  • 00:46:49
    but as the distance from the point
  • 00:46:51
    charge decreases
  • 00:46:52
    the electric field increases
  • 00:46:56
    there's two ways in which you can
  • 00:46:57
    increase the electric field
  • 00:46:59
    you can increase the magnitude of the
  • 00:47:01
    charge
  • 00:47:02
    which will cause the electric field to
  • 00:47:04
    go up
  • 00:47:05
    or
  • 00:47:06
    you could reduce the distance between
  • 00:47:08
    the point of interest and the charge
  • 00:47:11
    and that will also increase the electric
  • 00:47:13
    field
  • 00:47:18
    now let's move on to part c
  • 00:47:22
    the distance between the charge
  • 00:47:24
    and point x reduces by a factor of three
  • 00:47:29
    so what's going to be the the magnitude
  • 00:47:30
    of the new electric field in that case
  • 00:47:34
    so we're bringing it a lot closer to q
  • 00:47:37
    so here's the new
  • 00:47:39
    uh position of x
  • 00:47:42
    so here the distance doubled to 4 meters
  • 00:47:48
    but now it's going to be reduced by
  • 00:47:50
    factor three so it's two thirds
  • 00:47:53
    of a meter
  • 00:48:01
    so let's use this formula again
  • 00:48:04
    for part c k and q doesn't change
  • 00:48:08
    r is now one third of its original value
  • 00:48:12
    one times one is one one squared is one
  • 00:48:15
    three squared is nine
  • 00:48:17
    now we need to multiply the top and the
  • 00:48:18
    bottom by nine
  • 00:48:20
    one times nine is nine
  • 00:48:22
    one-ninth times nine the nines cancel we
  • 00:48:25
    get one
  • 00:48:26
    so the electric field is going to
  • 00:48:28
    increase by a factor of nine
  • 00:48:30
    a hundred times nine is nine hundred
  • 00:48:34
    so as you can see as we get closer
  • 00:48:36
    to the point charge
  • 00:48:38
    the magnitude of the electric field
  • 00:48:40
    greatly increases
  • 00:48:44
    now what happens if we triple the
  • 00:48:47
    magnitude of the charge
  • 00:48:49
    but at the same time reduce the distance
  • 00:48:51
    to one-fourth of its original value
  • 00:48:55
    so i won't draw a new picture for this
  • 00:49:00
    let's just get the answer
  • 00:49:03
    so k doesn't change
  • 00:49:05
    q
  • 00:49:06
    triples
  • 00:49:08
    and the distance is reduced to one
  • 00:49:09
    fourth of its value
  • 00:49:13
    so we have one times three which is
  • 00:49:14
    three
  • 00:49:15
    one squared is one four squared is
  • 00:49:17
    sixteen
  • 00:49:18
    so now i'm going to multiply the top and
  • 00:49:20
    the bottom by sixteen
  • 00:49:21
    so it's going to be three times sixteen
  • 00:49:23
    which is forty eight
  • 00:49:24
    so the electric field will increase by a
  • 00:49:27
    magnitude of forty eight
  • 00:49:29
    so the original electric field was a
  • 00:49:30
    hundred if we multiply that by 48
  • 00:49:34
    the new electric field will be 4 800
  • 00:49:37
    newtons per coulomb so this right here
  • 00:49:40
    is the answer to part d
  • 00:49:45
    so the reason why it's so high is
  • 00:49:46
    because
  • 00:49:48
    we've increased the charge
  • 00:49:50
    which causes e to go up and at the same
  • 00:49:53
    time we reduce
  • 00:49:55
    the distance which greatly
  • 00:49:57
    increase the value of e
  • 00:50:00
    so that's it for this problem
  • 00:50:04
    number eight
  • 00:50:05
    two identical point charges with a
  • 00:50:07
    magnitude of a hundred microclimbs
  • 00:50:10
    are separated by distance of one meter
  • 00:50:12
    as shown below
  • 00:50:14
    part a at what point will the net
  • 00:50:16
    electric field be equal to zero
  • 00:50:21
    will it be to the left will it be
  • 00:50:23
    between q1 and q2 or to the right
  • 00:50:26
    so let's identify three points of
  • 00:50:28
    interest
  • 00:50:31
    the first point will be somewhere to the
  • 00:50:33
    left which we'll call
  • 00:50:35
    point a
  • 00:50:37
    the second point will be somewhere in
  • 00:50:39
    the middle between q1 and q2 likely the
  • 00:50:41
    midpoint
  • 00:50:42
    and then c will be to the right
  • 00:50:46
    now to draw the electric field vector
  • 00:50:48
    created by q1
  • 00:50:50
    we need to draw a line from q1 to point
  • 00:50:52
    a
  • 00:50:53
    this is going to be e1
  • 00:50:55
    and for
  • 00:50:56
    e2 we're going to draw it from q2 to
  • 00:50:58
    point a
  • 00:51:00
    so at point a
  • 00:51:02
    both electric fields are moving to the
  • 00:51:05
    left
  • 00:51:06
    so there's going to be a net electric
  • 00:51:07
    field at point eight it's not going to
  • 00:51:09
    be zero but here's a question for you
  • 00:51:11
    which electric field will be greater e1
  • 00:51:12
    or e2
  • 00:51:14
    now remember
  • 00:51:16
    these charges are identical so the
  • 00:51:18
    magnitude of the charge is the same
  • 00:51:20
    the only thing that's different is the
  • 00:51:21
    distance
  • 00:51:23
    q1 is closer to point a than q2
  • 00:51:27
    so e1 is going to be bigger
  • 00:51:29
    than e2 so this is e1
  • 00:51:32
    e2 is going to be a smaller vector
  • 00:51:35
    nevertheless the net electric field at
  • 00:51:37
    point a will be directed west
  • 00:51:40
    now what about at point c
  • 00:51:43
    to draw e1 we're going to draw
  • 00:51:46
    a line from q1 to point c
  • 00:51:49
    and e2 will be from q2 to point c
  • 00:51:52
    now which one is bigger
  • 00:51:55
    q2 is closer to point c than q1 so e2 is
  • 00:51:58
    gonna be bigger at point two i mean e2
  • 00:52:01
    is going to be bigger at point
  • 00:52:03
    so for e1 we're going to draw a small
  • 00:52:05
    vector
  • 00:52:06
    and for e2 we're going to draw a bigger
  • 00:52:08
    vector
  • 00:52:10
    nevertheless the net electric field at
  • 00:52:12
    point c will be directed east
  • 00:52:15
    now what about at point b
  • 00:52:19
    e1 is going to be directed
  • 00:52:21
    away from q1 but towards point b
  • 00:52:24
    e2 will be directed away from q2
  • 00:52:28
    but towards point b
  • 00:52:31
    and q1 and q2 have the same charge and
  • 00:52:34
    at the midpoint at point b
  • 00:52:36
    they will be equally distant from point
  • 00:52:38
    b so if the charges are the same and the
  • 00:52:40
    distances are the same the magnitude of
  • 00:52:42
    e1 and e2 will be the same but they're
  • 00:52:44
    opposite in direction which means that
  • 00:52:46
    e1 and e2 cancels
  • 00:52:49
    so at point b the net electric field is
  • 00:52:52
    zero
  • 00:52:54
    so let's assume that q1 is the origin
  • 00:52:56
    it's at position zero
  • 00:52:58
    point b will be at 0.5 meters
  • 00:53:03
    so at 0.5 meters relative to the first
  • 00:53:06
    charge
  • 00:53:07
    then that electric field
  • 00:53:09
    will be zero e1 and e2
  • 00:53:12
    will cancel completely
  • 00:53:15
    so that's the answer for part a
  • 00:53:17
    at point b
  • 00:53:19
    or at the midpoint between q1 and q2 the
  • 00:53:21
    net electric field will be
  • 00:53:26
    zero now what about
  • 00:53:30
    part b
  • 00:53:36
    if the charge on q2
  • 00:53:39
    doubles
  • 00:53:40
    to 200 micro coulombs
  • 00:53:43
    where along the x-axis relative to the
  • 00:53:45
    first charge
  • 00:53:46
    will the net electric field be equal to
  • 00:53:48
    zero
  • 00:53:50
    and the distance between these two is
  • 00:53:51
    still the same
  • 00:53:56
    you think the net electric field will be
  • 00:53:58
    equal to zero between q1 and b or
  • 00:54:00
    between b and q2
  • 00:54:04
    now q2 is bigger
  • 00:54:07
    than q1
  • 00:54:09
    so at point b where they're equidistant
  • 00:54:12
    e1
  • 00:54:13
    is going to be a smaller vector
  • 00:54:15
    than e2 e2 is going to be twice as large
  • 00:54:19
    in order
  • 00:54:20
    to make these vectors equal we need to
  • 00:54:22
    increase e1 and decrease e2
  • 00:54:26
    if we can't change q1 and q2 the only
  • 00:54:28
    thing we could change is location
  • 00:54:30
    we need to move closer to q1 as we move
  • 00:54:33
    closer to q1 e1 is going to get bigger
  • 00:54:36
    e2 will get smaller and at some point
  • 00:54:38
    they're going to equal each other
  • 00:54:42
    so we're going to place point p
  • 00:54:44
    between q1 and b
  • 00:54:47
    somewhere between q1 and b
  • 00:54:49
    the net electric field will be equal to
  • 00:54:51
    zero and we need to find that point
  • 00:54:55
    so that point is going to be r1
  • 00:54:58
    r2
  • 00:55:00
    is the distance between q2 and point p
  • 00:55:05
    now let's call r1x
  • 00:55:07
    if r1 is equal to x what's r2
  • 00:55:11
    notice that the total distance is 1 so
  • 00:55:13
    r2 is going to be
  • 00:55:15
    1 minus x
  • 00:55:17
    if you add x and 1 minus x you're gonna
  • 00:55:20
    get one
  • 00:55:22
    so at this point
  • 00:55:23
    all we need to do to get the answer for
  • 00:55:26
    part b
  • 00:55:27
    is
  • 00:55:28
    get the value of x because x represents
  • 00:55:30
    the distance
  • 00:55:32
    relative to the first charge where the
  • 00:55:33
    net electric field will be equal to zero
  • 00:55:37
    so how can we calculate x
  • 00:55:40
    the net electric field at point b
  • 00:55:44
    remember at point b we have e1 which is
  • 00:55:46
    going towards the right and e2
  • 00:55:49
    i mean at point not b but point p
  • 00:55:52
    at point p we have e1 going to the right
  • 00:55:55
    and e2 is going to be going to the left
  • 00:56:00
    so because e1 is going to the right it's
  • 00:56:02
    positive e2 is going to the left it's
  • 00:56:04
    negative
  • 00:56:08
    now we want
  • 00:56:10
    point p is defined as the point where
  • 00:56:12
    the net electric field is zero so e1
  • 00:56:14
    minus e2 will be zero if we add e2 to
  • 00:56:17
    both sides
  • 00:56:18
    we'll get that e1 is equal to e2
  • 00:56:22
    they have to be the same in magnitude
  • 00:56:24
    but opposite direction
  • 00:56:26
    e1 is k
  • 00:56:28
    q1
  • 00:56:29
    over r1 squared
  • 00:56:31
    e2 is k
  • 00:56:33
    q2
  • 00:56:35
    over r2 squared
  • 00:56:38
    now let's divide both sides by k so we
  • 00:56:40
    can cancel
  • 00:56:41
    that
  • 00:56:42
    term let's replace r1 with x
  • 00:56:46
    so we have q1 over x squared and let's
  • 00:56:49
    replace r2
  • 00:56:50
    with one minus x
  • 00:56:53
    don't forget to square it
  • 00:57:00
    now let's cross multiply so here we're
  • 00:57:02
    going to have q 2
  • 00:57:04
    times x squared
  • 00:57:06
    and then this is going to be q 1 times
  • 00:57:09
    1 minus x squared
  • 00:57:12
    now let's replace q2 with
  • 00:57:14
    its value
  • 00:57:16
    and let's keep the unit micro coulombs
  • 00:57:19
    so we have 200 micro coulombs times x
  • 00:57:21
    squared
  • 00:57:22
    q1 is a hundred
  • 00:57:24
    micro coulombs
  • 00:57:27
    and what we're going to do at this point
  • 00:57:28
    is we're going to divide both sides by
  • 00:57:30
    100
  • 00:57:31
    micro coulombs
  • 00:57:34
    so the unit micro coulombs will cancel
  • 00:57:36
    on the left side
  • 00:57:38
    on the left we have two hundred over one
  • 00:57:40
    hundred which is two so we get two x
  • 00:57:42
    squared is equal to one
  • 00:57:45
    minus x squared
  • 00:57:54
    now we don't need to foil 1 minus x
  • 00:57:56
    squared
  • 00:57:58
    what we could do is take the square root
  • 00:57:59
    of both sides
  • 00:58:01
    so we're going to have the square root
  • 00:58:02
    of 2 and the square root of x squared is
  • 00:58:04
    just x
  • 00:58:06
    the square root of 1 minus x squared is
  • 00:58:08
    just 1 minus x
  • 00:58:11
    now the square root of 2 is 1.414
  • 00:58:15
    so now we need to do some algebra
  • 00:58:17
    let's add 1x to both sides
  • 00:58:21
    let me write that better
  • 00:58:24
    so there's a coefficient of one here
  • 00:58:27
    as we add one x to both sides
  • 00:58:33
    my handwriting is just not working today
  • 00:58:34
    i don't know why one point four one x
  • 00:58:37
    plus one x is two point four one x
  • 00:58:41
    and we can bring down the one on the
  • 00:58:42
    right side
  • 00:58:44
    so we have two point four one x
  • 00:58:46
    i mean two point four one four x is
  • 00:58:48
    equal to one so to get x by itself we
  • 00:58:51
    need to divide both sides
  • 00:58:53
    by
  • 00:58:54
    two point four one four
  • 00:58:59
    so x is going to be one
  • 00:59:02
    divided by two 2.414
  • 00:59:05
    and so we get 0.414
  • 00:59:10
    so at 0.414 meters
  • 00:59:14
    to the right of q1
  • 00:59:16
    which is point p at that point
  • 00:59:19
    the net electric field will be equal to
  • 00:59:21
    zero
  • 00:59:22
    so that's how we can calculate the exact
  • 00:59:24
    location
  • 00:59:26
    along the x axis
  • 00:59:28
    where the net electric field will be
  • 00:59:29
    equal to zero if q1 and q2 have
  • 00:59:34
    different magnitudes of charge
  • 00:59:58
    you
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