JEE MAIN - Physics (2002 - No. 47)
On moving a charge of $$20$$ coulomb by $$2$$ $$cm,$$ $$2$$ $$J$$ of work is done, then the potential differences between the points is
$$0.1$$ $$V$$
$$8$$ $$V$$
$$2V$$
$$0.5$$ $$V.$$
Explication
We know that $${{{W_{AB}}} \over q} = {V_B} - {V_A}$$
$$\therefore$$ $${V_B} - {V_A} = {{2J} \over {20C}} = 0.11J/C = 0.1V$$
$$\therefore$$ $${V_B} - {V_A} = {{2J} \over {20C}} = 0.11J/C = 0.1V$$