Two particles with charges +3.72 mu C and +1. | vedclass

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(B) Given,charge on the first particle,$Q_1 = +3.72 \mu C$ and charge on the second particle,$Q_2 = +1.86 \mu C$.The initial electrostatic force betwe

Vedclass · Accepted Answer

increases by $12 \%$

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(B) Given,charge on the first particle,$Q_1 = +3.72 \mu C$ and charge on the second particle,$Q_2 = +1.86 \mu C$.The initial electrostatic force between the charges is $F_1 = \frac{k Q_1 Q_2}{R^2} = \frac{k}{R^2} (3.72 	imes 1.86) 	imes 10^{-12} = \frac{k}{R^2} (6.9192 	imes 10^{-12}) \ N$.If $20 \%$ of $Q_1$ is transferred to $Q_2$,the new charges are:$Q_1^{\prime} = Q_1 - 0.20 Q_1 = 0.80 Q_1 = 0.80 	imes 3.72 = 2.976 \mu C$.$Q_2^{\prime} = Q_2 + 0.20 Q_1 = 1.86 + (0.20 	imes 3.72) = 1.86 + 0.744 = 2.604 \mu C$.The new electrostatic force is $F_2 = \frac{k Q_1^{\prime} Q_2^{\prime}}{R^2} = \frac{k}{R^2} (2.976 	imes 2.604) 	imes 10^{-12} = \frac{k}{R^2} (7.749504 	imes 10^{-12}) \ N$.The percentage change in force is $\frac{F_2 - F_1}{F_1} 	imes 100 = \frac{7.749504 - 6.9192}{6.9192} 	imes 100 = \frac{0.830304}{6.9192} 	imes 100 \approx 12 \%$.Since the value is positive,the force increases by $12 \%$. Thus,the correct option is $B$.

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

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