Question

a conducting sphere of radius 20 cm has a charge q placed on it producing a field of magnitude e0 at its surface. more charge is added to the surface until the magnitude of the field at the surface is 2 e0. how much charge was added to the surface?

Answer 1

To solve this problem, we can use the formula for the electric field due to a uniformly charged sphere:

E = (k * Q) / r^2,

Where:
E is the electric field,
k is the electrostatic constant (k = 9 × 10^9 N·m^2/C^2),
Q is the total charge of the sphere,
and r is the radius of the sphere.

Given that the original field magnitude is e0 and the final field magnitude is 2e0, we can set up the following equation:

e0 = (k * Q_initial) / (r^2),

2e0 = (k * Q_final) / (r^2).

By dividing these two equations, we can eliminate k and r^2:

(e0 / 2e0) = (Q_initial / Q_final).

Simplifying the equation gives us:

1/2 = Q_initial / Q_final.

Now, we can solve for Q_initial by substituting Q_final = q + Q_initial, where q is the charge added to the surface:

1/2 = Q_initial / (q + Q_initial).

Cross-multiplying, we get:

2 * Q_initial = q + Q_initial.

Simplifying this equation gives:

Q_initial = q.

Therefore, the amount of charge added to the surface is equal to the value of the initial charge, Q_initial, which is q.