A metal sphere of radius ra is supported on an insulating stand at the center of
ID: 1772548 • Letter: A
Question
A metal sphere of radius ra is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb. There is charge +q on the inner sphere and the charge -q on the outer spherical shell. (a) Calculate the potential V(r) for the following intervals. (Hint: The potential is the sum of potentials due to the individual spheres.) (Use any variable or symbol stated above along with the following as necessary: 0.) r < ra V = ra < r < rb V = r > rb V = (b) Show that the potential of the inner sphere with respect to the outer is egin{displaymath} V_{ab} = rac{q}{4 pi epsilon_0}( rac{1}{r_a} - rac{1}{r_b}) end{displaymath} (Do this on paper. Your instructor may ask you to turn in this work.) (c) Use Eq.(23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude egin{displaymath} E(r) = rac{V_{ab}}{( rac{1}{r_a} - rac{1}{r_b})} rac{1}{r^2} end{displaymath} (Do this on paper. Your instructor may ask you to turn in this work.) (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance r from the center, where r > ra. N/C
Explanation / Answer
given metal sphere, radius Ra
spherical shell, radius Rb
Charge on metal sphere = Q, charge on shell = -Q
a. for r < Ra
electric field inside conductors is 0
so E = 0
hence potential is constant inside the sphere and equal to the value at the surface of the sphere
so Va = kQ/Ra [ where k is coulombs constant]
for Ra < r < Rb
electric field, E
from gauss' law
E*4*pi*r^2 = Q/epsilon
E = kQ/r^2
hence dV = -Edr
integrating from r = Ra to r = r
V - Va = kQ[1/r - 1/Ra]
V - kQ/Ra = kQ[1/r] - kQ/Ra
V = kQ/r
for r > Rb
Electric field = E
from gauss Law
E*4*pi*r^2 = 0
E = 0
also, V = k(Q - Q)/r = 0
b. from previous part
potential difference between two spheres
dV = -kQdr/r^2
integrating from r = Ra to r = Rb
V = kQ(1/Ra - 1/Rb)
here k = 1/4*pi*epsilon
c. from part 1,. suing gasuu law
for Ra < r < Rb
E = kQ/r^2
d. for r > Ra
Qen = 0
so E = 0