Question

Two tiny spheres carrying charges $1.5\mu C$ and $2.5\mu C$ are located at $30cm$ apart. The potential at the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is

• A. $2.4\times {10}^{5}V$
• B. $4.2\times {10}^{3}V$
• C. $2.9\times {10}^{4}V$
• D. $3.7\times {10}^{5}V$

Answer 1

Answer: (A)

$2.4\times {10}^{5}V$

Here, $q_1=1.5\mu C=1.5\times {10}^{-6}C$

$q_2=2.5\mu C=2.5\times {10}^{-6}C$

Distance between the two spheres $=30cm=0.3cm$

(a) $V=?E=?$ For the mid point $0,r_1=r_2=\frac{0.3}{2}=0.15m$

$V=\frac{1}{4\pi {\in }_0[\frac{q_1}{r_1}+\frac{q_2}{r_2}]=\frac{9\times {10}^9(1.5+2.5)\times {10}^{-6}}{0.15}}$

$=2.4\times {10}^{5}volt$