# A question on electricity (potential difference)

1. Jul 21, 2010

### rkrthegreat

In the situation shown in the figure a particle having charge q is executing simple harmonic motion of amplitude A and time period T in front of a CONDUCTING EARTHED SPHERE of radius R. The reading of the voltmeter when the charge particle is at its mean position(at a distance r from the sphere's centre) is?

The attempt at a solution
I'm confused about the way to approach the problem. There could be so many ways in which the sphere could get a different potential, (like due to the electric field of the charge q, due to changing magnetic field because of the oscillating charge) that I am unable to get a pinpoint solution. Do help me with this one. Look at the attached diagram for clear details.

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2. Jul 21, 2010

### hikaru1221

Is there any other information, such as the resistance of the resistor or how fast the particle moves?

3. Jul 22, 2010

### rkrthegreat

Oh, I'm sorry, forgot to put that in the diagram. Take the resistance as R0. As for how fast the charge moves, the velocity at any position can be calculated, the amplitude of motion and time period being given in the question.

4. Jul 22, 2010

### hikaru1221

I mean, how great is T compared to some other characteristic times, such as r/c (c is speed of light), or the time for charge redistribution? You know, if T is very small compared to them, the problem will be greatly simplified.

5. Jul 23, 2010

### rkrthegreat

Yes, take T as very small

6. Jul 23, 2010

### hikaru1221

Oops :yuck: sorry. To simplify the problem, T should be very large, not very small. Apology :tongue: I guess you meant T was large, too?

If T is very large, we may turn the problem to a classical electrostatic one. The field formed by q is electrostatic field. Each position of q corresponds to an electrostatic image on the sphere, which comprises of 2 point charges: one at the center (q1), the other at somewhere midway between the center and q (q2). Denote V the potential of the sphere, then q1 is the one responsible for V: V=kq1/R^2, while q2 is for canceling the potential on the sphere by q, so q2 depends on x and q (I don't remember the formulas, but this can found in most books). Besides, the total charge on the sphere is q1+q2, and its time derivative is the current through the resistor, and thus: V = RoI = -Rod(q1+q2)/dt. You will arrive at a differential equation, and the rest is solving for V As I remember, the differential equation is quite crazy; but if A<<r, again, we can simplify the equation and it's solvable.

Last edited: Jul 23, 2010
7. Jul 24, 2010

### rkrthegreat

Thanks mate, I guess that's what was to be assumed, to reduce the question to a case of electrostatics. That was what I too needed to confirm, otherwise there could have been many paths to take on this question.
(I wrote take T as small because I thought you were proceeding to solve it using electromagnetics ;-) )