Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.
-What is the potential difference, V_B - V_A, between corners A and B?

Question

Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.

-What is the potential difference, V_B - V_A, between corners A and B?

Quizplus · Accepted Answer

The potential difference between two points in an electric field is given by the negative of the integral of the electric field along the path between the two points. In this case, the electric field is given by E = -kx, where k is a constant. The potential difference between A and B is then given by:V_B - V_A = -∫_A^B E dx = -∫₀^d (-kx) dx = -[-k(d²/2)] = kd²/2Substituting k = 1.2 × 10⁵ V/m and d = 1.4 × 10^-2 m, we get:V_B - V_A = (1.2 × 10⁵ V/m)(1.4 × 10^-2 m)² = -8.4 × 10⁵ VTherefore, the potential difference between A and B is -8.4 × 10⁵ V.

Two Point Charges Are Held at the Corners of a Rectangle