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Homework Help: Electrostatics surface charge Problems

  1. Mar 6, 2005 #1


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    Problem: Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral) conducting sphere of radius R. At the center of each cavity a point charge is placed -- call these carges [itex]q_a[/itex] and [itex]q_b[/itex].

    (a) Find the surface charges [itex]\sigma _a[/itex], [itex]\sigma _b[/itex], and [itex]\sigma _R[/itex].
    (b) What is the field outside the conductor?
    (c) What is the filed within each cavity?
    (d) What is the force on [itex]q_a[/itex] and [itex]q_b[/itex]?
    (e) Which of these answers would change if a third charge, [itex]q_c[/itex], were brought near the conductor?


    (a) I think [itex]q_a[/itex] would induce a charge of [itex]q_a[/itex] on the surface of the sphere, but this would include the surface of cavity b. At the same time, [itex]q_b[/itex] would induce a charge of [itex]-q_b[/itex] on the surface of its own cavity, so:

    [tex]\sigma _b = \frac{q_a - q_b}{4\pi b^2}[/tex]


    [tex]\sigma _a = \frac{q_b - q_a}{4\pi a^2}[/tex]


    [tex]\sigma _R = \frac{q_a + q_b}{4\pi R^2}[/tex]

    (b) [tex]\mathbf{E}(\mathbf{r}) = \frac{q_a + q_b}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]

    (c) We can treat cavity A as a spherical shell of radius a centered at the origin. The potential inside a spherical shell with uniform surface charge distribution is constant, so the electric field is zero due to the surface charge. Therefore, the only charge that matters is the point charge in the center. So the field inside is:

    [tex]\mathbf{E}(\mathbf{r}) = \frac{q_a}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]

    Except at the very center, since the field of the charge doesn't apply to itself, so the field there would be 0.

    (d) All effects that [itex]q_a[/itex] would have on [itex]q_b[/itex] are transmitted via the surface of cavity b, just as the effect on any particle outside the big sphere from the charges inside is "communicated" via the surface of the sphere. The position of the internal charges and the shapes of the cavities don't effect points outside the sphere, all that matters is the surface charge that ends up on the sphere. Similarly, the only thing that will affect [itex]q_b[/itex] is the surface it faces (and nothing beyond it), and since the surface around it is a sphere with a uniform charge distribution, and since [itex]q_b[/itex] is at the center of this sphere, the forces on it should cancel, and the net force on it should be 0; likewise for [itex]q_a[/itex].

    Also, since the field at the center is zero, the force is [itex]0 \times q_a = 0[/itex].

    (e) All except the last one.


    Have I done the above correctly? Our book talks a bit about how things work with one cavity, so the above answers are based on how I think those facts would extend to the case where there are 2 cavities, but I'm not entirely sure of the above, so please let me know, and if I've done anything wrong, let me know what's wrong about it, thanks.
  2. jcsd
  3. Mar 6, 2005 #2
    part a is not correct, otherwise great....

    use gauss law..... for [tex]\sigma_a [/tex] and [tex]\sigma_b[/tex]....you have their E field right in part c.
    Last edited: Mar 6, 2005
  4. Mar 6, 2005 #3


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    Thanks vincentchan. Gauss's Law gives:

    [tex]\oint _\mathcal{S} \mathbf{E}\cdot d\mathbf{a} = \frac{q_{enc}}{\epsilon _0}[/tex]

    But the field points in the same direction as the infinitessimal surface element, and it is constant on the surface, so:

    [tex]\frac{q_a}{4\pi \epsilon _0 a^2} (4\pi a^2) = \frac{q_a}{\epsilon _0}[/tex]

    This doesn't really get me anywhere. Where am I supposed to use [itex]\sigma _a[/itex] in Gauss's Law?
  5. Mar 7, 2005 #4


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    He means the field inside the conductor is 0, so what does Gauss's law tell you about the charge enclosed, say, in a surface right outside one of the shells?
  6. Mar 7, 2005 #5


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    Oh, I see. The enclosed charge must be zero, so:

    [tex]\sigma _a = -\frac{q_a}{4\pi a^2}[/tex]

    Thanks to both of you.
  7. Mar 7, 2005 #6


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    Oh wait. Since the charge enclosed inside a Gaussian surface enclosing one of the cavities must be zero, then even if a charge is brought near the surface of the sphere, the charge inside the Gaussian surface must remain zero, and so [itex]\sigma _a,\ \sigma _b[/itex] will not change. But wouldn't [itex]\sigma _R[/itex] change? The charge on the surface will not change, but there should be an induced negative (positive) distribution near the foreign charge, and thus a positive (negative) distribution on the opposite side of the sphere, so the distribution of the charge would change.

    Also, the fields inside the cavities should not change, and in fact, the only quantity computed in parts (a) through (d) that should change would be the field outside the big sphere and the charge distribution on the surface of the sphere.

    Is this right now?
  8. Mar 7, 2005 #7
    your post #5 is right... and your post #1 is right, too, except part a which you have already corrected...

    sure, it will change.... I thought you got it already....

  9. Mar 8, 2005 #8


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    I'm not understanding this problem clearly; so two cavities exists within the conductor with no particular orientation/position between the two right? I'm on this chapter myself so I would appreciate some help also.

    From the way I'm interpreting this problem...

    The sphere is in electrostatic equilibrium as a consequence from the disturbance caused by the two charges within each cavity of the conducting sphere. The surface boundary of cavity/conductor corresponding to each cavity has the charge induced by each of the point charges. Thus each of the inner surface of the conductor experiences a corresponding electric field; if the point charge is [itex]q+[/itex] within one of the cavities for instance, the charge induced within the inner surface of this cavitity would be [itex]q-[/itex].

    The charge of the sphere as a whole is neutral, thus the charge corresponding to the electric field at the outer edge of the conducting sphere is [itex]q_a+q_b[/itex]

    From this I do not understand how some of the answers obtained by AKG can be correct.

    For instance...

    Well, I thought that this can be looked at from the perpective of a gaussian surface with respect to a point charge, that is the electric field corresponding to columb's law, [itex]E=K_eq/r^2[/itex]. AKG seems to mention the equation relating to the electric field within an insulating sphere.

    But I'll wait for an critique before I go further.
  10. Mar 14, 2005 #9


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    You're both right. You just used [itex]K_e[/itex] where he used [itex]1/4\pi\epsilon_0[/itex], but these are the same value. That is [itex]\epsilon_0[/itex], the permittivity of free space, not the dielectric permittivity [itex]\epsilon[/itex].
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