An electric dipole is formed by two charges $+q$ and $-q$ located in $xy$-plane at $(0, 2) \mathrm{mm}$ and $(0, -2) \mathrm{mm}$, respectively, as shown in the figure. The electric potential at point $P(100, 100) \mathrm{mm}$ due to the dipole is $V_0$. The charges $+q$ and $–q$ are then moved to the points $(-1, 2) \mathrm{mm}$ and $(1, -2) \mathrm{mm}$, respectively. What is the value of electric potential at $P$ due to the new dipole?

• A

  $\frac{V_0}{4}$
• Correct answer
  B

  $\frac{V_0}{2}$
• C

  $\frac{V_0}{\sqrt{2}}$
• D

  $\frac{3V_0}{4}$

Hence, the correct answer is option: 'B'.