The potential difference $\left(V_A-V_B\right)$ in the arrangement shown in the figure is $(q=1 \mu \mathrm{C}, x=2 \mathrm{~cm}, y=3 \mathrm{~cm})$

• Correct answer
  A

  $5.4 \times 10^5 \mathrm{~V}$
• B

  $2.7 \times 10^5 \mathrm{~V}$
• C

  $5.4 \times 10^2 \mathrm{~V}$
• D

  $2.7 \times 10^2 \mathrm{~V}$

Solution

An arrangement is shown in the figure below,
Here, potential at a point $A$, $$ \begin{gathered} V_A=V_{A^{+q}}+V_{A^{-q}} \\ =\frac{9 \times 10^9 \times 10^{-6}}{2 \times 10^{-2}}-\frac{9 \times 10^9 \times 10^{-6}}{5 \times 10^{-2}}=9 \times 10^5\left[\frac{1}{2}-\frac{1}{5}\right] \end{gathered} $$ Similarly, potential at a point $B$, $$ V_B=V_{B^{+q}}+V_{B^{-q}}=9 \times 10^5\left[\frac{1}{5}-\frac{1}{2}\right] $$ Hence, $V_A-V_B=9 \times 10^5\left[\frac{1}{2}-\frac{1}{5}-\frac{1}{5}+\frac{1}{2}\right]$ $$ \Rightarrow \quad V_A-V_B=5.4 \times 10^5 \mathrm{~V} $$ Hence, the correct option is (a).

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