A metal sphere with radius 1.00 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.40 cm. Charge +q is put on the inner sphere and charge −q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 470 V, with the inner sphere at higher potential.

Calculate q.

q = 5.84*10^-10 C

Explanation:

Given

Radius of the metal sphere, r(a) = 1 cm = 1*10^-2 m

Radius of the spherical shell, r(b) = 9.4 cm = 9.4*10^-2 m

Potential difference between the spheres, ΔV = 470 V

The potential difference of an inner sphere is given by

V(in) = kq/r(a) - kq/r(b)

while the potential difference of the outer sphere is given as

V(out) = 0

ΔV = V(in) - V(out)

ΔV = kq/r(a) - kq(b)

ΔV = kq * [1/r(a) - 1/r(b)]

where k = 9*10^9

470 = 9*10^9 * q (1/0.01 - 1/0.094)

470 = 9*10^9 * q *(100 - 10.638)

470 = 9*10^9 * q * 89.362

q = 470 / (9*10^9 * 89.362)

q = 470 / 8.04*10^11

q = 5.84*10^-10 C

dang that balls went far