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(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.Point $A$ is at co

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(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.Point $A$ is at coordinates $(0, a)$,so its distance from the origin $O$ is $r_A = \sqrt{0^2 + a^2} = a$.Point $B$ is at coordinates $(a, 0)$,so its distance from the origin $O$ is $r_B = \sqrt{a^2 + 0^2} = a$.The potential at $A$ is $V_A = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.The potential at $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.Since $V_A = V_B$,the potential difference $\Delta V = V_B - V_A = 0$.The work done $W$ in moving a charge $-Q$ from $A$ to $B$ is given by $W = (-Q) \cdot \Delta V$.Substituting $\Delta V = 0$,we get $W = (-Q) \cdot 0 = 0$.

As per the diagram,a point charge $+q$ is placed at the origin $O$. The work done in taking another point charge $-Q$ from point $A$ [coordinates $(0, a)$] to point $B$ [coordinates $(a, 0)$] along the straight path $AB$ is:

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