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(B) The electric potential $V$ due to a point charge $+q$ at a distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r}$.At point $A(

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(B) The electric potential $V$ due to a point charge $+q$ at a distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r}$.At point $A(0, a)$,the distance from the origin is $r_A = a$. Thus,the potential is $V_A = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a}$.At point $B(a, 0)$,the distance from the origin is $r_B = a$. Thus,the potential is $V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a}$.The work done $W$ in moving a charge $-Q$ from $A$ to $B$ is given by $W = (-Q)(V_B - V_A)$.Since $V_A = V_B$,the potential difference $(V_B - V_A) = 0$.Therefore,the work done $W = (-Q)(0) = 0$.

$A$ point charge $+q$ is placed at the origin. The work done in moving another point charge $-Q$ along the straight line path $AB$ from point $A$ with coordinates $(0, a)$ to point $B$ with coordinates $(a, 0)$ is .......

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