Physics - Electrostatics Question with Solution | TestHub

Let two charges are$q_1\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d} q_2$and r is the distance between them, then electrical force

$F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_1{q}_2}{r^2}=200\mathrm{ }\mathrm{N}$..... (i)

If$q_1$is increased by 10%, then

$q_1^{\prime }=\frac{110}{100}{q}_1$

and$q_2$is decreased by 10%, then

$q_2^{\prime }=\frac{90}{100}{q}_2$

Then, the electrical force between them,

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1^{\prime }{q}_2^{\prime }}{r^2}$

$F^{\prime }=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{110}{100}{q}_1\times \frac{90}{100}{q}_2}{r^2}$.... (ii)

From equations (i) and (ii), we get

$F^{\prime }=200\times \frac{99}{100}\Rightarrow F^{\prime }=198\mathrm{ }\mathrm{N}$