1. Initial configuration

The charges are at:

• $+q$ at $(0,2)\,\text{mm}$
• $-q$ at $(0,-2)\,\text{mm}$

Point of observation is $P(100,100)\,\text{mm}.$

Since the distance of $P$ from the dipole is much larger than the dipole size, we use the dipole potential formula:

$V = \frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot \vec r}{r^3}$

where $\vec p$ is the dipole moment and $\vec r$ is the position vector of the field point from the dipole center.

2. Dipole moment in the initial case

The center of the dipole is at the origin. The dipole moment points from $-q$ to $+q$, i.e. upward along $+y$.

Separation vector magnitude: $d = 4\,\text{mm}$

So, $\vec p_1 = q(4\,\hat j)\,\text{mm}$

Thus, $\vec p_1 = (0,4q) \quad \text{(in mm units)}$

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

Therefore, $\vec p_1\cdot \vec r = 0\cdot 100 + 4q\cdot 100 = 400q$

Hence, $V_0 \propto 400q$

(the factor $\frac{1}{4\pi\varepsilon_0 r^3}$ is same for comparison).

3. New configuration

Now the charges are moved to:

• $+q$ at $(-1,2)\,\text{mm}$
• $-q$ at $(1,-2)\,\text{mm}$

The center is still at the origin, because midpoint is $\left(\frac{-1+1}{2},\frac{2+(-2)}{2}\right)=(0,0).$

Dipole moment points from $-q$ to $+q$: $\vec p_2 = q[(-1,2)-(1,-2)] = q(-2,4)$

So, $\vec p_2 = (-2q,4q)$

Now, $\vec p_2\cdot \vec r = (-2q)(100) + (4q)(100) = -200q+400q=200q$

Thus new potential $V' \propto 200q$

4. Ratio of new potential to old potential

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

So, $V' = \frac{V_0}{2}$

5. Option check

• A: $\frac{V_0}{4}$ ❌
• B: $\frac{V_0}{2}$ ✅
• C: $\frac{V_0}{\sqrt{2}}$ ❌
• D: $\frac{3V_0}{4}$ ❌

Therefore, the correct answer is B.