Question

Two point charges are separated by a distance d. The first has a charge of +2e, and the second has a charge of +3e. How does the electric potential energy of the +2e charge change if it is moved to a new position, so that it is separated from the second charge by a distance of 2d? Use k = 9.00 × 10⁹ N⋅m²/C² for Coulomb's constant.

a) It increases by a factor of 2.
b) It increases by a factor of 4.
c) It decreases by a factor of 2.
d) It decreases by a factor of 4.

Answer 1

The electric potential energy of the +2e charge decreases by a factor of 2 when it is moved from a distance d to 2d away from the +3e charge.

Step-by-step explanation:

The electric potential energy of the +2e charge when moved to a new position at a distance of 2d from the +3e charge can be determined using the formula for electric potential energy between two point charges, which is:

U = (k * Q1 * Q2) / r, where U is the potential energy, k is Coulomb's constant, Q1 and Q2 are the point charges, and r is the separation distance between the charges.

Initially, the potential energy (U1) is given by:

U1 = (k * 2e * 3e) / d

When the charge is moved to twice the distance, the new potential energy (U2) is:

U2 = (k * 2e * 3e) / (2d)

U2 = (1/2) * (k * 2e * 3e) / d = U1 / 2

This calculation shows that the potential energy decreases by a factor of 2. Therefore, the correct answer to the question is (c) It decreases by a factor of 2.