Question

Two charges of +10 $\mu$C and +20 $\mu$C are separated by a distance of $2cm$. The net potential (electric) due to the pair at the middle point of the line joining the two charges, is :

• A. $27MV$
• B. $18MV$
• C. $20MV$
• D. $23MV$

Answer 1

The correct option is A $27MV$

$\text{Step 1: Potential due to each charge at mid point (O)}$

Potential due to $q_1$ at $O:$ $V_1=\frac{1}{4\pi {\in }_0}\frac{q_1}{r_1}=\frac{9\times {10}^9\times 10\times {10}^{-6}C}{(1\times {10}^{-2}m)}$$=9\times {10}^{6}V=9MV$

Potential due to $q_2$ at $O:$ $V_2=\frac{1}{4\pi {\in }_0}\frac{q_2}{r_2}=\frac{9\times {10}^9\times 20\times {10}^{-6}C}{(1\times {10}^{-2}m)}$$=18\times {10}^{6}V=18MV$

$\text{Step 2: Net potential at the middle point}$

Net potential: $V=V_1+V_2$

$=9MV+18MV=27MV$

Hence, Option(A) is correct.