1)An insulating sphere of radius a=4 cm has charge Q=10 uC uniformly distributed

Question

1)An insulating sphere of radius a=4 cm has charge Q=10 uC uniformly distributed throughout.This sphere is surrounded by two concentric conducting shells.The first has an inner rdius of b=7.5cm and an outer radius of c=8.0cm.  The seocnd has an inner radius of d=13cm and an outer radius of e=15cm.  (GIVE UNTS IN ALL ANSWERS)

A)Using Gauss's Law , derive the Electric Field at r=3cm?

B)What is the net electric flux through a cubic surface with sides of 9.0cm centered about the insulating sphere?

C)What is the Electric field at (i) r=10 cm; (ii)r=14cm and r=15 cm?

D)What, if any, is the total surface charge density on the inner surface of the outer conducting shell?

E)If excess negative charge is placed onto the smaller of the two conducting shells so that the electric field for r>15 cm is zero, how large must that charge density be?

Explanation / Answer

Let a = 4 cm, b = 7.5 cm and c = 8 cm. also let q =10 uC. Now the charge density in the conductiong sphere is

p = 3q/(4*pi*a^3)

Consider 0 < r < a. Guass' law gives

Integral(E*dA) = q'/eo =1/e0 * integral(p*dV) = 4*pi*p/e0 * integral(r^2dr) = 4*pi*r^3*p/(3*e0) = q/e0*(r/a)^3

Integral(E*dA) = E*4*pi*r^2 = q/e0 * (r/a)^3 ---> E = q*r/(4*pi*e0*a^3)


Now consider a < r < b and apply Gauss' law:

Integral (E*dA) = q/e0 ---> E*4*pi*r^2 = q/e0 ----> E = q/(4*pi*e0*r^2)

When r = b, E = E(b) = q/(4*pi*e0*b^2)

For b< r < c, you are inside the conductor and the field is by defnition 0 ----> E = 0

For r > c Gauss's law gives

E*4*pi*r^2 = (q+q1)/e0 where q1 = -1 uC

E = (q+q1)/(4*pi*e0*r^2)

You can plug the numbers and plot the E field versus radius-------ask me if you need any help

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