An electric dipole is formed by two charges +q and −q located in the xy-plane at (0, 2) mm and (0, −2) mm, respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V₀. The charges +q and −q are then moved to the points (−1, 2) mm and (1, −2) mm, respectively. What is the value of the electric potential at P due to the new dipole?

Correct Answer: $\frac{V_0}{2}$

Step 1: Understand the formula for electric potential due to a short electric dipole
The electric potential $V$ at a far point $P$ (with position vector $\overrightarrow{r}$) due to an electric dipole having dipole moment vector $\overrightarrow{p}$ located at the origin is given by:

$V=\frac{1}{4\pi {\epsilon }_0}\frac{\overrightarrow{p}·\overrightarrow{r}}{r^3}$

Since point $P(100,100)$ mm is at a much larger distance compared to the dimensions of the dipole (a few millimeters), we can treat it as a short electric dipole centered near the origin.

Step 2: Calculate the initial dipole moment and potential $V_0$
Initially, charge $+q$ is at $(0,2)$ mm and charge $-q$ is at $(0,-2)$ mm.
The position vector pointing from $-q$ to $+q$ is:

$\overrightarrow{d}=(0-0)\hat{i}+(2-(-2\left)\right)\hat{j}=4\hat{j}\text{ mm}$

Thus, the initial dipole moment vector is:

${\overrightarrow{p}}_1=q\overrightarrow{d}=4q\hat{j}$

The position vector of point $P$ relative to the origin is:

$\overrightarrow{r}=100\hat{i}+100\hat{j}\text{ mm}$

Taking the dot product ${\overrightarrow{p}}_1·\overrightarrow{r}$:

${\overrightarrow{p}}_1·\overrightarrow{r}=\left(4q\hat{j}\right)·(100\hat{i}+100\hat{j})=400q$

Therefore, the initial potential $V_0$ is proportional to $400q$:

$V_0=k·\left(400q\right)$ (where $k=\frac{1}{4\pi {\epsilon }_0{r}^3}$)

Step 3: Calculate the new dipole moment and potential $V^{\text{'}}$
As shown in the figure, the charges are moved such that $+q$ is at $(-1,2)$ mm and $-q$ is at $(1,-2)$ mm.
The new displacement vector from $-q$ to $+q$ is:

${\overrightarrow{d}}^{\text{'}}=(-1-1)\hat{i}+(2-(-2\left)\right)\hat{j}=-2\hat{i}+4\hat{j}\text{ mm}$

The new dipole moment vector is:

${\overrightarrow{p}}_2=q{\overrightarrow{d}}^{\text{'}}=q(-2\hat{i}+4\hat{j})$

Taking the dot product ${\overrightarrow{p}}_2·\overrightarrow{r}$:

${\overrightarrow{p}}_2·\overrightarrow{r}=q(-2\hat{i}+4\hat{j})·(100\hat{i}+100\hat{j})=q(-200+400)=200q$

The new potential $V^{\text{'}}$ at point $P$ is:

$V^{\text{'}}=k·\left(200q\right)$

Step 4: Relate the new potential to $V_0$
Comparing $V^{\text{'}}$ with $V_0$:

$\frac{V^{\text{'}}}{V_0}=\frac{200q}{400q}=\frac{1}{2}$

$V^{\text{'}}=\frac{V_0}{2}$

Hence, the value of the electric potential at point $P$ due to the new dipole is $\frac{V_0}{2}$.