1. A nonconducting shell of radii a and 2a has a uniform charge density p. Pleas

Question

1. A nonconducting shell of radii a and 2a has a uniform charge density p.

Please answer parts (a) through (c) and show all work.

AP PHYSICS C TEST SECTION II-ELECTRICITY AND MAGNETISM Time: 45 minutes 3 Questions Directions: Answer all 3 questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. (l A nonconducting shell of radii a and 2a has a uniform charge density (a) Determine the electric field in the following regions I. (i) r> 2a (ii) 2a > r> a (b) Determine the electric potential in the regions (i) r> 2a (ii) 2a > > a iii) r

Explanation / Answer

The shell has uniform charge density of p, with the inner and outer radii being a and 2a

Part a.) For calculating the electric field at various points, we will consider a spherical Gaussian surface which concentric with the given shell and then we will use the Gauss's law to find the electric field at the respective points.

i.) Let us consider Gaussian surface or r > 2a. By Gauss's law we get:

E(4r2) = Charge enclosed / o

or, E(4r2) = 4p[8a3 - a3]/3o = 28pa3/3o

or, E = 7pa3/3r2o is the required electric field.

ii.) For 2a> r > a

we can write: E(4r2) = 4p[8r3 - a3]/3o

or, E = p[8r3 - a3]/3r2o is the required expression.

ii.) For r<a, there is no charge enclosed, hence electric field would be zero.

Part b.) (i.) We know that the electric potential is given as:

dv/dr = E = 7pa3/3r2o

or, dv = 7pa3/3r2o dr

Also, we assume that the potential at infinity is zero, hence we get:

V = 7pa3/3ro

ii.) For a < r < 2a

we have: Vs - Vr = p[8r3 - a3]/3r2o dr

or, Vr = 7pa2/6o - p8r2/6o - pa3/3ro is the expression for electric potential.

iii.) For any point r < a

The electric field inside is zero in this case, hence the potential at the inner surface of the shell would also be the potential for the any point inside r < a

From the previous we get:

Vr = 7pa2/6o - p8a2/6o - pa2/3o = 2pa2/3o is the required expression for electric potential for any point where r < a

Part c.) When the non-conductor is covered by conductors, the electrons on the conductor re arrange themselves in such a manner that the electric field inside the body of the conductor remains zero.

That would mean the potential inside the body conductor would remain constant, hence no change would in potential would happen from r = 4a to r = 2a, hence the net potential inside the non-conductor would drop.

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