Question

Two tiny spheres carrying charges $1.8\mu C$ and $2.8\mu C$ are located at $40cm$ 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

Answer: (B)

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 $O$,
$V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]$
$=\frac{9\times 10^9\left(1.8+2.8\right)\times 10^{-6}}{0.2}$
$=2.1\times 10^{5}V$