Question

A capacitor is composed of two cylindrical conducting
shells. The inner shell has a radius A, is centered inside
the outer shell and has a positive surface charge density
+3s. The outer shell has radius B = 3A and negative
surface charge density -s. Assume the length of the
conductors is ?infinitely? long compared to the radius B
so that you can ignore all edge effects. Let r be the
vector pointing from the center of the capacitor to any

1. Find the E field in the regions r < A , A < r < B and r > B (20 pts)
2. Find the potential difference DVAB between r = A and r = B. (10 pts)
3. Find the capacitance per unit length of the configuration. (10 pts)
4. Find the energy per unit length stored in the capacitor. (5 pts)
point in space.

A capacitor is composed of two cylindrical conducting shells. The inner shell has a radius A, is centered inside the outer shell and has a positive surface charge density +3o. The outer shell has radius B 3A and negative surface charge density -o. Assume the length of the conductors is "infinitely" long compared to the radius B so that you can ignore all edge effects. Let r be the vector pointing from the center of the capacitor to any point in space. 1. Find the E field in the regions r <A , A < r < B and B (20 pts) 2. Find the potential difference AVAB betweenr A and r B. (10 pts) 3. Find the capacitance per unit length of the configuration. (10 pts) 4. Find the energy per unit length stored in the capacitor. (5 pts)

Answer 1

For r < A we cut a Gaussian surface inside the inner cylinder, because there is no any charge inside the inner surface so, net charge inside the Gaussian surface = q_eu = 0 Now, by gauss's law, countourintegral E^Vector . ds^vector = q_eu/epsilon_0 = 0/epsilon_0 E. 2 pi r l = 0/epsilon_0 E = 0 For A < r < B, Because surface charge density + 3 sigma for inner cylinder so, net charge inside the Gaussian surface = q_eu = 3 sigma 2 pi A l Now, By gauss's law countourintegral E^vector. ds^vector = q_eu/epsilon_0 E. 2 pi r l = 3 sigma 2 pi A l/epsilon_0 E = 3 sigma A/epsilon_0 r r^N/c r^= unitvector that is along r. \r\n for r > B, Net charge inside the Gaussian surface = + 3 sigma 2 pi A l - sigma 2 pi B l = 6 sigma pi A l - 6 sigma pi A l q_eu