A neutral conducting spherical shell has inner and outer radii r1 = 8 cm and r2 = 10 cm. A charge q = 6.1 µC is placed at the center of the sphere.

Question

A neutral conducting spherical shell has inner and outer radii r₁ = 8 cm and r₂ = 10 cm. A charge q = 6.1 µC is placed at the center of the sphere. What is the charge density of the inner surface (r₁) of the shell?

Answer 1

All charge on a conductor distributes evenly around the outer surfaces. The conductor has no electric field within it because integral E * da = Qenclosed/ epsilon naught would have a Qenc = 0, causing E to = 0. In general, a conductor cannot have an electric field within it because the charge enclosed is 0, no matter what. The conductor will create this scenario even when you place a charge at the center.

When positive charge q is placed at the center, the inner shell will distribute a charge of -q around its surface so that the Q enclosed between the distances r1 < d < r2 = 0. Imagine the gaussian dotted line surface to be somewhere between r1 and r2; then, the Q enclosed is +q + -q = 0.

Charge on inner shell = -q = -6.1 * 10^-6 C

The charge density of the inner surface would be sigma = charge / surface area

Charge = -6.1 * 10^-6 C

Surface Area sphere = 4*pi*r^2 (use r1 = 0.08 m for this inner surface) = 4*pi*(0.08)^2 = 0.08 m^2

Surface charge density sigma = -6.1*10^-6 / 0.08

= -7.58 * 10^-5 C/m^2