Question

The electric potential inside a charged spherical conductor of radius R is given by V = keQ/R, and the potential outside is given by V = keQ/r. Using Er = -dV/dr, derive the electric field inside and outside this charge distribution. (Use any variable or symbol stated above as necessary.)

Answer 1

Answer 2

a) E = -dV/dr = -d(keQ/R) /dr = 0

b) E = -dV/dr = -d (keQ/r) dr = -[-keQ/r^2] = keQ/r^2

Answer 3

Let the density of the charge inside the sphere be q [C/m³].

A. According to Gauss' Law, the flux of the vector E through a sphere with radius r and concentric with the given sphere, will be:

1) For r E*4pr² = (4/3)pr³q/e0 ==> E= qr/(3e0)

2) For r=R:
E*4pr² = (4/3)pR³q/e0 ==> E=R³q/(3e0r²)

B. The potential will be

1) For r f=?[r,8] E dr = ?[r,R] qt/(3e0) dt + ?[R,8] R³q/(3e0r²) dr =
qt²/(6e0) [t=r to R] - R³q/(3e0r) [r=R to 8] =
qR²/(6e0) - qr²/(6e0) + R²q/(3e0) = qR²/(2e0) - qr²/(6e0)

2) For r=R:
f=?[r,8] E dr = ?[r,8] R³q/(3e0t²) dt = -qR³/(3e0t) [t=r to 8] = qR³/(3e0r)