When an electric charge of 0.05 C passes in one conductor, 100 mJ work is done, and in the other

When an electric charge of 0.05 C passes in one conductor, 100 mJ work is done, and in the other, when a charge of 80 mC flows, 0.5 J work. On which conductor is the voltage higher and how many times?

To find the voltage ratio on the indicated conductors, we will use the formula: k = U1 / U2 = (A1 / q1) / (A2 / q2) = A1 * q2 / (A2 * q1).
Values of variables: A1 – work of current in the first specified conductor (A1 = 100 mJ = 0.1 J); q2 – second charge (q2 = 80 mC = 0.08 C); A2 – work of current in another conductor (A2 = 0.5 J); q1 is the first charge (q1 = 0.05 C).
Calculation: k = A1 * q2 / (A2 * q1) = 0.1 * 0.08 / (0.5 * 0.05) = 0.32 or U2 / U1 = 3.125 p.
Answer: The voltage on the second indicated conductor is 3.125 times greater.