Question

Two point charges $A=+3 \, nC$ and $B=+1 \, nC \, $ are placed $5 \, cm$ apart in the air. The work done to move charge $B$ towards $A$ by $1 \, cm$ 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,
$A=+ \, 3 \, nC=3\times 10^{- 9 \, }C$
$B=+ \, 1 \, nC=1\times 10^{- 9 \, }C$
Distance, $r_{1}=5 \, cm=0.05 \, m=5\times 10^{- 2} \, m$
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, $r_{2}=r_{1}-1$
$r_{2}=5-1=4 \, cm=0.04 \, m=4\times 10^{- 2} \, m$
$=kq_{1}q_{2}\left[\frac{1}{r_{2}} - \frac{1}{r_{1}}\right]$
$=9\times 10^{9}\times 3\times 10^{- 9}\times 1\times 10^{- 9}$
$\left[\frac{1}{4 \times 10^{- 2}} - \frac{1}{5 \times 10^{- 2}}\right]$
$=\frac{27 \times 10^{- 9} \times 1}{5 \times 4 \times 10^{- 2}}$
$=\frac{27}{20}\times 10^{- 7}$
$=1.35\times 10^{- 7}J$