An electric dipole is formed by two charges +𝑞 and −𝑞 located in 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 electric potential at P due to the new dipole? (A) 𝑉0/4 (B) 𝑉0/2 (C) 𝑉0/√2 (D) 3𝑉0/4

To determine the electric potential at point $P$ due to the new dipole configuration, we need to understand how the electric potential of a dipole behaves and how it changes with the orientation and position of the dipole.

Step 1: Understanding the Electric Potential of a Dipole

The electric potential $V$ at a point due to a dipole is given by:

$$V=\frac{1}{4\pi {ϵ}_0}\frac{\mathbf{p}\cdot \mathbf{r}}{r^3}$$

where:

• $\mathbf{p}$ is the dipole moment vector,

• $\mathbf{r}$ is the position vector from the dipole to the point where the potential is being calculated,

• $r$ is the magnitude of $\mathbf{r}$,

• ${ϵ}_0$ is the permittivity of free space.

The dipole moment $\mathbf{p}$ is defined as:

$$\mathbf{p}=q\mathbf{d}$$

where $q$ is the charge and $\mathbf{d}$ is the displacement vector from the negative charge to the positive charge.

Step 2: Initial Configuration

In the initial configuration:

• The positive charge $+q$ is at $(0,2)$ mm.

• The negative charge $-q$ is at $(0,-2)$ mm.

The dipole moment ${\mathbf{p}}_1$ is:

$${\mathbf{p}}_1=q{\mathbf{d}}_1=q((0,2)-(0,-2))=q(0,4)\text{ mm}$$

The position vector $\mathbf{r}$ from the dipole to point $P$ at $(100,100)$ mm is:

$$\mathbf{r}=(100,100)\text{ mm}$$

The electric potential $V_0$ at point $P$ due to the initial dipole is:

$$V_0=\frac{1}{4\pi {ϵ}_0}\frac{{\mathbf{p}}_1\cdot \mathbf{r}}{r^3}$$

Step 3: New Configuration

In the new configuration:

• The positive charge $+q$ is moved to $(-1,2)$ mm.

• The negative charge $-q$ is moved to $(1,-2)$ mm.

The new dipole moment ${\mathbf{p}}_2$ is:

$${\mathbf{p}}_2=q{\mathbf{d}}_2=q((-1,2)-(1,-2))=q(-2,4)\text{ mm}$$

The position vector $\mathbf{r}$ remains the same:

$$\mathbf{r}=(100,100)\text{ mm}$$

The electric potential $V$ at point $P$ due to the new dipole is:

$$V=\frac{1}{4\pi {ϵ}_0}\frac{{\mathbf{p}}_2\cdot \mathbf{r}}{r^3}$$

Step 4: Comparing the Potentials

To find the ratio $\frac{V}{V_0}$, we compare the dot products ${\mathbf{p}}_1\cdot \mathbf{r}$ and ${\mathbf{p}}_2\cdot \mathbf{r}$:

$${\mathbf{p}}_1\cdot \mathbf{r}=(0,4)\cdot (100,100)=0\times 100+4\times 100=400$$$${\mathbf{p}}_2\cdot \mathbf{r}=(-2,4)\cdot (100,100)=-2\times 100+4\times 100=-200+400=200$$

Thus, the ratio is:

$$\frac{V}{V_0}=\frac{{\mathbf{p}}_2\cdot \mathbf{r}}{{\mathbf{p}}_1\cdot \mathbf{r}}=\frac{200}{400}=\frac{1}{2}$$

Therefore, the electric potential at point $P$ due to the new dipole is:

$$V=\frac{V_0}{2}$$

Final Answer

(B) $V_0\mathrm{/}2$