Two particles with charges \(+3.72 \mu \mathrm{C}\) and \(+1.86 \mu \mathrm{C}\) are some distance apart. If…
Two particles with charges \(+3.72 \mu \mathrm{C}\) and \(+1.86 \mu \mathrm{C}\) are some distance apart. If \(20 \%\) of the charge is transferred from first particle to second particle then the electrostatic force between them is
decreases by \(12 \%\)
increases by \(12 \%\)
increase by \(4 \%\)
decreases by \(4 \%\)
Solution
Given, charge on the first particle, \(Q_1=+3.72 \mu \mathrm{C}\) and charge on second particle, \(Q_2=1.86 \mu \mathrm{C}\)
Then the electrostatic force between charges,
\(\begin{aligned}
F_1 & =\frac{k Q_1 Q_2}{R^2} \\
& =\frac{k}{R^2}(3.72 \times 1.86) 10^{-12} \mathrm{~N} \\
F_1 & =\frac{k}{R^2}\left(6.9192 \times 10^{-12}\right)
\end{aligned}\)
If \(20 \%\) of \(Q_1\) is given to \(Q_2\),
\(\begin{aligned}
& Q_2^{\prime}=Q_2+Q_1 \times \frac{20}{100}=1.86+3.72 \times 0.20 \\
& \Rightarrow \quad Q_2^{\prime}=2 \cdot 604 \mu \mathrm{C} \\
& \text {and } \quad Q_1^{\prime}=Q_1 \times \frac{80}{100}=2.976 \mu \mathrm{C}
\end{aligned}\)
Hence,
\(\begin{aligned}
& F_2=\frac{k}{R^2} Q_1^{\prime} \cdot Q_2^{\prime}=\frac{k}{R^2} 2.976 \times 2.604 \times 10^{-12} \\
& F_2=\frac{k}{R^2} 7.7495
\end{aligned}\)
So, % increment in \(F_2\),
\(\begin{aligned}
\frac{F_2-F_1}{F_1} \times 100 & =\frac{k}{R^2} \frac{(7.7495-6.9192) \times 10^{-12}}{\frac{k}{R^2} \times 6.9192 \times 10^{-12}} \times 100 \\
& =12 \%
\end{aligned}\)
Hence, the correct option is (b).