Question

When a $2\mu C$ charge is carried from point A to point B, the amount of work done by the electric field is $50\mu J$. What is the potential difference and which point is at a higher potential?
(A) 25 V, B
(B) 25 V, A
(C) 20 V, B
(D) Both are at the same potential

Answer 1

We are given that a charge is moved from one point to another. We know if work is done in moving a charge from one point to another then there exists a potential difference between those two points. We know that the relationship between work done, charge and potential is given by: $W=Vq$, where W is the work done and V is the potential.

Complete step by step answer:
Given charge, $q=2\mu C=2\times {{10}^{-6}}C$
Work done, $W=50\mu J=50\times {{10}^{-6}}J$
So, the potential can be given as: $V=\dfrac{W}{q}$
$\Rightarrow V=\dfrac{W}{q}$
$\Rightarrow V=\dfrac{50\times {{10}^{-6}}}{2\times {{10}^{-6}}}$
$\Rightarrow V=25V$
So, the potential difference comes out to be 25 Volts.
But the charge was moved from point A to B, thus
$\Rightarrow V={{V}_{A}}-{{V}_{B}}$
$\therefore {{V}_{A}}-{{V}_{B}}=25V$

Thus A is at higher potential. So, the correct option is B.

Note: Electric potential is a scalar quantity and it is assumed to be zero at the infinity. If we want to move a charge from one point to another and if the electric potential is same at both the points work done will be zero because no work is required to be done to move a charge on an equipotential surface.The electric potential is related to electric field and electric field is defined as the negative gradient of electric potential.