Question

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential difference? Recall that Fe = qE.

Answer 1

The electric potential difference is 7.5 V.

Electric potential difference between two points is defined as the work done in moving unit positive charge between the two points. Electric potential difference can also be defined as the work done W per unit charge q in an electric field.

$\Delta V=\frac{W}{q}$......(1)

Work done by a force is the product of the force F and the displacement s made in the direction of the force. Assuming that the displacement is parallel to the line of action of the force,

$W=F.s$......(2)

Rewrite equation (1) using equation (2).

$\Delta V=\frac{Fs}{q}$......(3)

Substitute the given values of force, displacement and charge in the equation (3).

$\Delta V=\frac{Fs}{q}\\ =\frac{(25*10^-^6N)(15m)}{(50*10^-^6C)} \\ =7.5V$

The potential difference between the two points is 7.5 V