PQRS is square of side $1 \mathrm{~m}$. A charge of $100 \mu \mathrm{C}$ is placed at the centre of the square. Then the work done to take $3 \mu \mathrm{C}$ charge from the corner $\mathrm{P}$ to the corner $\mathrm{R}$.

The electric potential $V$ at any point at a distance $r$ from a point charge $q$ is given by $V = \dfrac{kq}{r}$.

Let the charge at the centre of the square be $Q = 100 \mu \text{C} = 100 \times 10^{-6} \text{C}$.

The distance from the centre of a square of side $a = 1 \text{m}$ to any of its corners is $r = \dfrac{a}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \text{m}$.

Since the distance from the centre to corner P is $r_P = \dfrac{1}{\sqrt{2}} \text{m}$ and the distance from the centre to corner R is $r_R = \dfrac{1}{\sqrt{2}} \text{m}$, the potential at P and R are equal:

$V_P = \dfrac{kQ}{r_P} = \sqrt{2}kQ$

$V_R = \dfrac{kQ}{r_R} = \sqrt{2}kQ$

The work done $W$ in moving a charge $q_0 = 3 \mu \text{C}$ from P to R is given by $W = q_0(V_R - V_P)$.

Since $V_R = V_P$, the potential difference $V_R - V_P = 0$.

Therefore, $W = q_0 \times 0 = 0 \text{J}$.

Answer: Zero