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Spherical conductor with cavities

by meteorologist1
Tags: cavities, conductor, spherical
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meteorologist1
#1
Jan17-05, 09:47 PM
P: 101
Hi all, I need help on the following problem:

A solid spherical conductor S contains 2 small spherical cavities (not-concentric). The total charge on the conductor is zero, but at the center of each cavity there are two point charges q1 and q2 respectively. At large distance r away from the center of the sphere, there is a third charge Q. What force exists on each of the four objects in this problem: Q, S, q1, and q2? Which answers, if any, are approximate and depend on r being relatively large?

Thanks.
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learningphysics
#2
Jan18-05, 12:41 AM
HW Helper
P: 4,125
Quote Quote by meteorologist1
Hi all, I need help on the following problem:

A solid spherical conductor S contains 2 small spherical cavities (not-concentric). The total charge on the conductor is zero, but at the center of each cavity there are two point charges q1 and q2 respectively. At large distance r away from the center of the sphere, there is a third charge Q. What force exists on each of the four objects in this problem: Q, S, q1, and q2? Which answers, if any, are approximate and depend on r being relatively large?

Thanks.
For the problem I'll call the S system: S, q1, q2.

Can you use Gauss' law to find the electric field due to the S system (the whole thing cavities charges and all) at a distance r? Use this to get the force on Q.

What is the electric field in a cavity in a conductor? This should give you the force on q1 and q2.

Finally, by newton's third law, I'd say that the force on Q by the S system=force on the S system by Q. So this gives the total force acting on S,q1,q2 together. Subtract the force acting on q1, and the force acting on q2, and you get the force acting on S alone.
meteorologist1
#3
Jan18-05, 08:48 AM
P: 101
Ok, so by Gauss's Law I would have E = (q1+q2)/(4pi epsilon r^2). So the force on Q would be F = QE.

The electric field in q1's cavity would be E = -q1/(4pi epsilon s^2), where s is the radius of the cavity? I think that a charge of -q1 would be induced on the inner wall of the conductor, but I'm not sure. Similarly for q2.

Please correct me if there's a problem. Thanks.

learningphysics
#4
Jan18-05, 11:43 AM
HW Helper
P: 4,125
Spherical conductor with cavities

Quote Quote by meteorologist1
Ok, so by Gauss's Law I would have E = (q1+q2)/(4pi epsilon r^2). So the force on Q would be F = QE.

The electric field in q1's cavity would be E = -q1/(4pi epsilon s^2), where s is the radius of the cavity? I think that a charge of -q1 would be induced on the inner wall of the conductor, but I'm not sure. Similarly for q2.

Please correct me if there's a problem. Thanks.
You've got the force on Q right.

You've got the right idea for the electric field in the cavities. But what you need is the electric field created by all the charges except the one in the cavity. I apologize, I didn't mention this in my previous post. You get the total electric field created by all charges except q1. Multiply by q1, to get the force on q1.

When you have a cavity in a conductor, all charges outside the cavity create a total of 0 electric field inside the cavity. Your textbook might have a proof of this.

So force on q1=0. force on q2=0.

The force on S=(force on Q)-0-0=force on Q (which you already know).


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