Question

Two point charges 3*10^-6 and 8*10^-6 repel each other by a force of 6*10^-3n. if each of them is given an additional charge -6* 10^6 c

Answer 1

Answer:

so, where is the question.

plz check your question and post it again

Answer 2

Answer:

The correct answer is $1.5 \times 10^{-3} \mathrm{~N}$(attractive)

Explanation:

The electrostatic force between the charges is given by,

$\begin{aligned}&F=\frac{k Q_{1} Q_{2}}{r^{2}} \\&\therefore F \propto Q_{1} Q_{2} \\&\Rightarrow \frac{F_{1}}{F_{2}}=\frac{Q_{1} Q_{2}}{Q_{1}^{\prime} Q_{2}^{1}} \\&=\frac{3 \times 10^{-6} \times 8 \times 10^{-6}}{\left(3 \times 10^{-6}-6 \times 10^{-6}\right)\left(8 \times 10^{-6}-6 \times 10^{-6}\right)}=\frac{3 \times 8}{-3 \times 2}=-\frac{4}{1} \\&\Rightarrow F_{2}=-\frac{F_{1}}{4}=-\frac{6 \times 10^{-3}}{4}=-1.5 \times 10^{-3} \mathrm{~N}\end{aligned}$

Since now $Q_{1}^{\prime}$ is negative and $Q_{2}^{\prime}$ positive therefore the electrostatic force between them becomes attractive.