Two point charges are initially separated by a distance of in vacuum

PHYSICS · Multi-part · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

Two point charges are initially separated by a distance of 0.20 m0.20\,\text{m} in vacuum. Charge 1 is +2.0×10−6 C+2.0\times10^{-6}\,\text{C} and charge 2 is −4.0×10−6 C-4.0\times10^{-6}\,\text{C}. The charges are then slowly moved so that their final separation is 0.50 m0.50\,\text{m} while an external agent prevents any change in their kinetic energies. (a) Calculate the initial electric potential energy UiU_i of the system. (b) Calculate the final electric potential energy UfU_f of the system. (c) Determine the work done by the external agent during the process. State whether this work is positive or negative and justify your answer. (d) If the external agent were removed and the charges were allowed to move from 0.50 m0.50\,\text{m} back to 0.20 m0.20\,\text{m}, describe the sign of the work done by the electric force and interpret its physical meaning for the motion of the charges.

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