In the figure below, a solid sphere, of radius a = 2.20 cm is concentric with a
ID: 1696905 • Letter: I
Question
In the figure below, a solid sphere, of radius a = 2.20 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net charge q1 = +1.00 fC which is distributed uniformly through the sphere; the shell has a net charge of q2 = -q1.
In the figure below, a solid sphere, of radius a = 2.20 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net charge q1 = +1.00 fC which is distributed uniformly through the sphere; the shell has a net charge of q2 = -q1. (a) What is the magnitude of the electric field at radial distance r = 0? N/C (b) What is the magnitude of the electric field at radial distance r = a/2? N/C (c) What is the magnitude of the electric field at radial distance r = a? N/C (d) What is the magnitude of the electric field at radial distance r = 1.30a? N/C (e) What is the magnitude of the electric field at radial distance r = 2.10a? N/C (f) What is the magnitude of the electric field at radial distance r = 3.00a? N/C (g) What is the net charge on the inner surface of the shell? fC (h) What is the net charge on the outer surface of the shell? fCExplanation / Answer
The shell is made from conducting metal. Use a concentric spherical Gaussian surface which has the radius of 2.2a. (inside the shell). Because this place is inside the metal shell so there is no electric field. Gauss's law. E*S=q_total*epsilon0. so q_total*epsilon0=0. so q_total=0. so the charge on the inner surface of the shell is -q1=q2. So there is no charge on the outer surface of the shell ( all the charge distributed on the inner surface). ------------------- a)r=0. E=0 ( this area is inside both the sphere and the shell). b)r=a/2. E=0 (same as above) c)r=a. E=kq1/a^2=0.0186(N/C) (out side of the sphere but still inside the shell). d)r=1.3a. E=kq1/(1.3a)^2=0.011(N/C) (same as above) e)r=2.1a and r=3a, use gauss'law here with the gaussian surface be a concentric spherical one with radius of 2.1a and 3a respectively. In both situation, total q is zero (look at reasoning above). so E=0. f)inner. q=q2=-1fC. g)outer. q=0fC