For the configuration shown in Figure P24.50, suppose that a = 5.0 cm, b = 20 cm
ID: 1525406 • Letter: F
Question
For the configuration shown in Figure P24.50, suppose that a = 5.0 cm, b = 20 cm, and c = 25 cm. Furthermore, suppose that the electric field at a point r_1 = 10 cm from the center is measured to be E_1 = 3100 N/C radially inward, while the electric field at a point r_2 = 50 cm from the center is E_2 = 190 N/C radially outward. From this information, Find the charge (with sign) on the insulating sphere. Find the charge (with sign) on the inner surface of the hollow sphere. Find the charge (with sign) on the outer surface of the hollow sphere. If the inner charged sphere could be removed, what charge would remain on the hollow conducting sphere?Explanation / Answer
(A) From Gauss Law,
total flux = E . A = Qin / e0
for Gaussian sphere surface of r = 0.10 m
Qin = Qa
E ( 4 pi r^2) = Qa / e0
(3100) (4 x pi x 0.10^2) = Qa / (8.854 x 10^-12)
Qa = 3.45 x 10^-9
as field is inwards that means Qa is negative hence
Qa = - 3.45 x 10^-9 C Or - 3.45 nC
(B) field inside the conductor is always zer0.
for b < r < c , E= 0 hence Qin = 0
Qin = Qa + Qb = 0
Qb = - Qa = 3.45 nC
(C) for r > c
Qin = Qa + Qb + Qc and Qa + Qb = 0
Qin = Qc
(190) (4 x pi x 0.50^2) = Qc / (8.854 x 10^-12)
Qc = 5.28 x 10^-9 C
as field is radially outward hence Qc is +ve.
Qc = + 5.28 nC
(D) charge will not change.
Qnet = Qb + Qc = 8.73 nC