Question Detail

Given:Electric field (E) on the surface = 2.9 x 104 N/CRadius of the sphere (R) = 6 cm = 0.06 mAccording to Gauss's Law, for a spherically symmetric charge distribution or a hollow sphere with charge enclosed, the electric field outside (or on the surface) is given by:E = kQ / R2Where k = 1 / (4πε₀) = 9 x 109 Nm2/C2.We need to find the enclosed charge (Q):Q = E * R2 / kQ = (2.9 x 104 N/C) * (0.06 m)2 / (9 x 109 Nm2/C2)Q = (2.9 x 104) * (0.0036) / (9 x 109)Q = (0.1044 x 104) / (9 x 109)Q = 1044 / (9 x 109) = 116 x 10-9 C = 0.116 x 10-6 C = 0.116 µC.Let's re-examine the options: 0.012 µC is significantly different. Let's re-calculate using 0.012µC to see if it matches the given E. If Q = 0.012 µC = 0.012 x 10-6 C.E = (9 x 109) * (0.012 x 10-6) / (0.06)2E = (0.108 x 103) / 0.0036 = 108 / 0.0036 = 30000 N/C = 3.0 x 104 N/C.This value (3.0 x 104 N/C) is very close to the given electric field (2.9 x 104 N/C). The discrepancy likely arises from rounding in the question's values or options.Therefore, 0.012 µC is the closest and most probable answer.