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(B) The electric potential due to a dipole at a point defined by position vector $\vec{r}$ is given by $V = \frac{k \vec{p} \cdot \vec{r}}{r^3}$.For t

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$V_0 / 2$

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(B) The electric potential due to a dipole at a point defined by position vector $\vec{r}$ is given by $V = \frac{k \vec{p} \cdot \vec{r}}{r^3}$.For the initial dipole,the dipole moment is $\vec{p}_1 = q(4 \hat{j}) 	ext{ mm} = 4q \hat{j} 	ext{ mm}$.The position vector of point $P$ is $\vec{r} = 100 \hat{i} + 100 \hat{j} 	ext{ mm}$.Thus,$V_0 = \frac{k (4q \hat{j}) \cdot (100 \hat{i} + 100 \hat{j})}{r^3} = \frac{k (400q)}{r^3}$.For the new dipole,the charges are at $(-1, 2)$ and $(1, -2)$. The dipole moment vector $\vec{p}_2$ is directed from $-q$ to $+q$:$\vec{p}_2 = q [(-1 - 1) \hat{i} + (2 - (-2)) \hat{j}] = q (-2 \hat{i} + 4 \hat{j}) 	ext{ mm}$.The potential at $P$ due to the new dipole is $V = \frac{k \vec{p}_2 \cdot \vec{r}}{r^3}$.$V = \frac{k [q (-2 \hat{i} + 4 \hat{j})] \cdot (100 \hat{i} + 100 \hat{j})}{r^3} = \frac{k q (-200 + 400)}{r^3} = \frac{k (200q)}{r^3}$.Comparing $V$ and $V_0$:$V = \frac{k (200q)}{r^3} = \frac{1}{2} \left( \frac{k (400q)}{r^3} ight) = \frac{V_0}{2}$.

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