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 $Q_{2}$,
$Q_{2}'=Q_{2}+Q_{1} \times \frac{20}{100}=1.86+3.72 \times 0.20$
$\Rightarrow \,Q_{2}'=2 \cdot 604\, \mu C$
and $Q_{1}' =Q_{1} \times \frac{80}{100}=2.976 \,\mu C$
Hence, $ F_{2} =\frac{k}{R^{2}} Q_{1}' \cdot Q_{2}'=\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} $
$\frac{F_{2}-F_{1}}{F_{1}} \times 100= \frac{k}{R^{2}} \frac{(7.7495-6.9192) \times 10^{-12}}{\frac{k}{R^{2}} \times 6.9192 \times 10^{-12}} \times 100$
$=12 \%$