Question

Two point charges $A=+3nC$ and $B=+1nC$ are placed $5cm$ apart in the air. The work done to move charge $B$ towards $A$ by $1cm$ is

• A. $2.0\times 10^{-7}J$
• B. $2.7\times 10^{-7}J$
• C. $12.1\times 10^{-7}J$
• D. $1.35\times 10^{-7}J$

Answer 1

Answer: (D)

Given,
$A=+3nC=3\times 10^{-9}C$
$B=+1nC=1\times 10^{-9}C$
Distance, $r_1=5cm=0.05m=5\times 10^{-2}m$
Work done $\left(W\right)=U_B-U_A$
$=\frac{k{q}_1{q}_2}{r_2}-\frac{k{q}_1{q}_2}{r_1}$
Here, $r_2=r_1-1$
$r_2=5-1=4cm=0.04m=4\times 10^{-2}m$
$=k{q}_1{q}_2\left[\frac{1}{r_2}-\frac{1}{r_1}\right]$
$=9\times 10^9\times 3\times 10^{-9}\times 1\times 10^{-9}$
$\left[\frac{1}{4\times 10^{-2}}-\frac{1}{5\times 10^{-2}}\right]$
$=\frac{27\times 10^{-9}\times 1}{5\times 4\times 10^{-2}}$
$=\frac{27}{20}\times 10^{-7}$
$=1.35\times 10^{-7}J$