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

Answer: (B)

$2.1 \times 10^{5} \, V$

Answer 2

Here, $q_{1}=1.8\,\mu C=1.8\times10^{-6}\,C$, $q_{2} = 2.8 \mu C = 2.8 \times 10^{-6}C$ Distance between the two spheres $= 40 \,cm = 0.4\, m$ For the mid-point $r_{1}=r_{2}=\frac{0.40}{2}=0.2\,m$ Potential at $O$, $V=\frac{1}{4\pi\varepsilon_{0}} \left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}\right]$ $=\frac{9\times10^{9}\left(1.8+2.8\right)\times10^{-6}}{0.2}$ $=2.1\times10^{5}\, V$