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A hollow conducing sphere of inner radius R is centered at the origin. The sphere is split into two pieces with a voltage given by

$V=V_0, 0 \leq \theta \leq \theta_0$

$V=-V_0, \theta_0 < \theta < 2\pi$

A point charge is located at the origin. Find the force on the charge.

My Attempt:

I face a conundrum. By Gauss' Law, the electric field E inside the sphere is 0 and by $F=qE+qv\times B$ the force should be 0.

However, the voltage is not symmetrical, and therefore there is a net potential energy, and therefore there should be a force. If my reasoning is correct the force would be something like $2V_0 qR$ in the direction of $(0,\theta_0/2,0)$. But I am very iffy on my rational if that is the case.

Update:

It occured to me to try using $E=-\nabla V$, but considering $\nabla V$, wouldn't $\frac{\partial V}{\partial \rho}= V_0\delta(\rho-R)$ for $0\leq \theta \leq\theta_0$, and $\frac{\partial V}{\partial \rho}= -V_0\delta(rho-R)$ elsewhere? by the same rational, $\frac{\partial V}{\partial \theta}= 2V_0\delta(\theta-\theta_0)$. If that is the case I have no idea how to determine the force.

Update: I think I worked it out, It's $sV_0*q/d$ where q is the total charge on the sphere * 2 * $\theta_0/2\pi$. Does that seem right?

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  • $\begingroup$ Gauss's Law does not predict the the electric field is zero. (and take a closer look at your "something like" force. Close, but ...) $\endgroup$
    – garyp
    Commented Apr 26, 2018 at 20:13
  • $\begingroup$ Ok good to know. My problem then arises from simply not knowing the relationship between potential and force. I should think it would be fairly simple, as with the case of gravitational potential and force, but I'm having a lot of trouble with it. $\endgroup$
    – Seven
    Commented Apr 27, 2018 at 18:54
  • $\begingroup$ For that, see this SE question $\endgroup$
    – garyp
    Commented Apr 27, 2018 at 19:08

2 Answers 2

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By Gauss' Law, the electric field E inside the sphere is 0

How do you reason?

When we use Gauss' law to show that the E field inside a uniformly charged spherical shell is 0, we rely on the symmetry of the problem to tell us the flux density through the surface must be uniform. Without that symmetry, Gauss's law, in the integral form, can't tell you that the field inside the sphere is 0.

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  • $\begingroup$ I was figuring that because the flux is proportional to the charge inside the closed surface, and there was no charge inside, that the flux must be 0. Since posting I realized my obvious mistake, that there is a charge inside the surface. I was disregarding it because I was thinking of it as a "test charge," which I now realize was erroneous. $\endgroup$
    – Seven
    Commented Apr 27, 2018 at 17:30
  • $\begingroup$ The test charge is taken to be vanishingly small, and does not contribute to the field being tested. Furthermore, the field produced by a charge does not contribute a force on the charge. In other words, ignore the field produced by the test charge. $\endgroup$
    – garyp
    Commented Apr 27, 2018 at 18:09
  • $\begingroup$ @Seven, The net flux through the surface is 0 when there's no charge enclosed. But current can be going "in" one part of the surface and "out" another part of the surface. Sometimes (but not in your example) we can use symmetry to show that the flux must be equal over all parts of the surface (so it must all be in or all out). Or that it must be 0 on some part of the surface, and use that to reason about the other parts of the surface. $\endgroup$
    – The Photon
    Commented Apr 27, 2018 at 19:29
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The two hemispheres must be separated by a narrow circular gap. All of the charge is concentrated there and a strong electric field exists in the gap. These two circles of charge, in the limit of an very thin gap, cause no net electric field in the origin.

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  • $\begingroup$ They aren't necessarily hemispheres. $\theta_0$ is not necessarily $\pi$, but some arbitrary angle. But I take your meaning. All that being said, how do I determine the charge in the gap using only the Voltage? $\endgroup$
    – Seven
    Commented Apr 27, 2018 at 17:34
  • $\begingroup$ I don't believe that this is true. In the limit you propose, the separating ring will be populated by point dipoles, and the field at the center would not be zero. $\endgroup$
    – garyp
    Commented Apr 27, 2018 at 18:00

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