Question

A point charge

q = −4.0 ✕ 10−12 C
is placed at the center of a spherical conducting shell of inner radius 3.4 cm and outer radius 3.9 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 7.5 N/C.
a) What is the charge density (in C/m2) on the inner surface of the shell?
b) What is the charge density (in C/m2) on the outer surface of the shell?
c) What is the net charge (in C) on the conductor?

Answer 1

To find the charge density on the inner and outer surfaces of the conductor shell, we use the equation q = σ(4πR²). For the net charge on the conductor, we simply sum the charges on the inner and outer surfaces.

Step-by-step explanation:

To find the charge density on the inner surface of the shell, we can use the equation q = σ(4πR²). Rearranging the equation, we have σ = q/(4πR²), where q is the charge on the inner surface of the shell and R is the radius of the inner surface. Plugging in the values q = -4.0 x 10⁻¹² C and R = 3.4 cm (or 0.034 m), we can calculate the charge density σ.

For the outer surface of the shell, we use the same equation with the outer radius. Plugging in q = -4.0 x 10⁻¹² C and R = 3.9 cm (or 0.039 m), we can calculate the charge density σ.

The net charge on the conductor is the sum of the charges on the inner and outer surfaces. So the net charge Q on the conductor is Q = q_inner + q_outer. Plugging in the values q_inner = -4.0 x 10⁻¹² C and q_outer = 4.0 x 10⁻¹² C, we can calculate the net charge Q.