When a 3 µC of charge is carried from point A to point B, the amount of work done byelectric field is 75 µJ. Determine the potential difference?

Correct option is B

​$$\textbf{Given:} \\\text{Charge, } Q = 3\,\mu C \\\text{Work done, } W = 75\,\mu J \\[10pt]\textbf{We know, }\text{The potential difference } (\Delta V) \text{ between two points is given by:} \\\Delta V = \frac{W}{Q} \\[10pt]\Delta V = \frac{75\,\mu J}{3\,\mu C} = 25\,V \\[10pt]\text{So, the potential difference between points A and B is 25 volts.}$$​​