Question

Two tiny spheres carrying charges 1.8 $\mu C$ and $2.8\mu C$ are located at 40 cm apart. The potential at the mid-point of the line joining the two charges is

• A. $3.8\times {10}^4$ V
• B. $2.1\times {10}^5$ V
• C. $4.3\times {10}^4$ V
• D. $3.6\times {10}^5$ V

Answer 1

The correct option is B $2.1\times {10}^5$ V

Here $q_1=1.8\mu C=1.8\times {10}^{-6}C,q_2=2.8\mu C=2.8\times {10}^{-6}C$
Distance between the two spheres $=40cm=0.4m$
For the mid-point $r_1=r_2=\frac{0.40}{2}=0.2m$
Potential at $O$
$V=V_1+V_2=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{r_1}+\frac{q_2}{r_2}]=\frac{9\times {10}^9(1.8+2.8)\times {10}^{-6}}{0.2}=2.1\times {10}^{5}V$