The electric charge moves from a point with a potential of 125 V to a point with a potential of 75 V.

The electric charge moves from a point with a potential of 125 V to a point with a potential of 75 V. In this case, the forces of the electrostatic field do the work of 1 mJ.

Task data: φ1 (field potential at the starting point) = 125 V; φ2 (field potential at the end point) = 75 V; A (work done by the forces of the electrostatic field) = 1 mJ (1 * 10-3 J).

We express the magnitude of the electric charge from the formula: A = q * (φ1 – φ2), whence q = A / (φ1 – φ2).

Calculation: q = 1 * 10-3 / (125 – 75) = 0.02 * 10-3 C = 20 μC.

Answer: The amount of charge is 20 μC.