A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm,outer radius = 2.0 cm)which has a net charge of -4.0 nC.Determine the resulting charge density on the inner surface of the conducting sphere.

Question

Quizplus · Accepted Answer

The charge inside the conductor is distributed uniformly throughout the surface of the sphere. Therefore, the net charge on the inner surface of the conductor is equal to the charge of the point charge.

The charge density (ρ) on the inner surface of the conductor is given by:

ρ = Q/A

where Q is the charge on the inner surface of the conductor and A is the area of the inner surface of the conductor.

The area of the inner surface of the conductor is given by:

A = 4πr^2

where r is the radius of the inner surface of the conductor.

In this case, r = 1.0 cm = 0.01 m

Therefore,

A = 4π(0.01)^2 = 1.26 x 10^-3 m^2

The charge on the inner surface of the conductor is equal to the charge of the point charge:

Q = 6.0 nC = 6.0 x 10^-9 C

Therefore,

ρ = Q/A = (6.0 x 10^-9 C)/(1.26 x 10^-3 m^2) = 4.8 x 10^-3 C/m^2

This is equivalent to 4.8 μC/m^2.

The correct answer is B.

A Point Charge of 6 μ\muμ C/m2 B)-4 μ\muμ C/m2 C)-9 μ\muμ C/m2 D)+9

A Point Charge of 6 $\mu$ C/m²
B)-4 $\mu$ C/m²
C)-9 $\mu$ C/m²
D)+9