Question

A charge of 240C is moved when energy of 45J is applied between two points. Find the voltage between the two points.

Answer 1

$E= 45 \;J \\ Q= 240 \;C \\ E= \frac{1}{2}CV^2 \\ Q=CV \\ E= \frac{1}{2}(CV)V \\ E= \frac{1}{2}QV \\ V= \frac{2E}{Q} \\ V= \frac{2 \times 45}{240}= 0.375 \;Volt$