A spherical conducting shell has a uniform surface charge density of 1.5 $\times$ 10^-7 C/m².What is its radius if the electric field 1.5 m from its center is 17 N/C outward? A) 3.9 cm B) 4.7 cm C) 3.2 cm D) 9.5 cm E) 6.4 cm

Question

Quizplus · Accepted Answer

For a conducting shell, the electric field at any point outside the shell is the same as if all the charge is concentrated at the center of the sphere. So we can use Coulomb's Law to find the total charge Q enclosed within a sphere of radius r:

k(Q)/(r^2)=(17 N/C)x(4 π (1.5 m)^2)

Simplifying and solving for Q:

Q = (17 N/C)x(4 π (1.5 m)^2) x (r^2)/k

The surface charge density is given by:

σ = Q/(4πr^2)

Substituting Q into the equation above and solving for r:

r = √(Q/(4πσ))

r = √(((17 N/C)x(4 π (1.5 m)^2) x (r^2)/k)/(4π(1.5 x 10^-7 C/m^2)))

r = √(4.08 x r^2)

r = 4.7 cm (to two significant figures)

A Spherical Conducting Shell Has a Uniform Surface Charge Density ×\times×

A Spherical Conducting Shell Has a Uniform Surface Charge Density $\times$