The electric field just outside the surface of a hollow conducting sphere of radius 20 cm has a magnitude of 500 N/C and is directed outward.An unknown charge Q is introduced into the centre of the sphere and it is noted that the electric field is still directed outward but has decreased to 100 N/C.What is the magnitude of the charge Q?

Question

Quizplus · Accepted Answer

Since the electric field just outside the surface of the hollow conducting sphere is directed outward, we know that there must be a negative charge on the surface of the sphere that is attracting positive charges towards it, generating the outward electric field.

When the unknown charge Q is introduced at the center of the sphere, there is now a positive charge present that generates its own electric field. This field adds to the field generated by the negative charges on the surface of the sphere, resulting in a smaller net field just outside the surface.

We can use Gauss's law to relate the net charge encapsulated by the sphere to the electric field just outside its surface:

Φ_E = q_enc / ε0 = E * A

where Φ_E is the electric flux, q_enc is the net charge encapsulated by the sphere, ε0 is the permittivity of free space, E is the electric field just outside the surface of the sphere, and A is the surface area of the sphere.

Since the sphere is hollow, we know that there is no net charge inside the sphere – the positive charge Q introduced to the center of the sphere is exactly cancelled by negative charges on the surface of the sphere. Therefore, the net charge encapsulated by the sphere is just equal to the negative charge on its surface.

At the initial state, with a field of 500 N/C, we have:

500 N/C * 4π(0.20 m)^2  = -q / ε0

Solving for q, we get:

q = -1.13 nC

At the final state, with a field of 100 N/C, we have:

100 N/C * 4π(0.20 m)^2 = -q / ε0

Solving for q, we get:

q = -0.22 nC

The increase in charge from -1.13 nC to -0.22 nC corresponds to a charge of:

Δq = -0.22 nC - (-1.13 nC) = 0.91 nC

Since the unknown charge Q is positive, we must have:

Q = -Δq = 1.8 nC

Therefore, the answer is B.

The Electric Field Just Outside the Surface of a Hollow