Electric Charge and Field - AP Physics 2 Exam Question

A small sphere with a charge of +2 μC is placed at point A, as shown in the diagram below. The sphere is in a region where there is a uniform electric field of magnitude 200 N/C directed toward the right. The sphere experiences an electrostatic force of 0.4 N toward the left.

1. Determine the magnitude and direction of the force exerted on an electron placed at point B, which is 2 meters to the left of point A.

Answer:

Given: Charge of the sphere, q = +2 μC = 2 × 10^-6 C Electric field, E = 200 N/C Electrostatic force on the sphere, F = 0.4 N

The force experienced by the sphere is given by the equation:

F = qE

Since the force is opposite to the direction of the electric field, the force and the electric field have opposite signs. Hence, the force on an electron placed at point B is given by:

F = -eE

where e is the charge of an electron, e = -1.6 × 10^-19 C.

Substituting the given values:

-1.6 × 10^-19 C = (-1.6 × 10^-19 C)(200 N/C)

Simplifying, we find:

F = 1.6 × 10^-19 C × 200 N/C

F = 3.2 × 10^-17 N

Therefore, the magnitude of the force exerted on an electron placed at point B is 3.2 × 10^-17 N, and the direction of the force is toward the right.