Question

Two particles with charges $+3.72\mu C$ and $+1.86\mu C$ are some distance apart. If $20\%$ of the charge is transferred from first particle to second particle then the electrostatic force between them is

• A. decreases by $12\%$
• B. increases by $12\%$
• C. increases by $4\%$
• D. decreases by $4\%$

Answer 1

Answer: (B)

Given, charge on the first particle,
$Q_1=+3.72\mu C$ and charge on second particle,
$Q_2=1.86\mu C$
Then the electrostatic force between charges,
$F_1=\frac{k{Q}_1{Q}_2}{R^2}$
$=\frac{k}{R^2}(3.72\times 1.86)10^{-12}N$
$F_1=\frac{k}{R^2}\left(6.9192\times 10^{-12}\right)$
If $20\%$ of $Q_1$ is given to $Q_2$,
$Q_2^{\prime }=Q_2+Q_1\times \frac{20}{100}=1.86+3.72\times 0.20$
$\Rightarrow Q_2^{\prime }=2\cdot 604\mu C$
and $Q_1^{\prime }=Q_1\times \frac{80}{100}=2.976\mu C$
Hence, $F_2=\frac{k}{R^2}{Q}_1^{\prime }\cdot Q_2^{\prime }=\frac{k}{R^2}2.976\times 2.604\times 10^{-12}$
$F_2=\frac{k}{R^2}7.7495$
So, $\%$ increment in $F_2$
$\frac{F_2-F_1}{F_1}\times 100=\frac{k}{R^2}\frac{(7.7495-6.9192)\times 10^{-12}}{\frac{k}{R^2}\times 6.9192\times 10^{-12}}\times 100$
$=12\%$