Two charges $2 \mathrm{C}$ and $6 \mathrm{C}$ are separated by a finite distance. If a charge of $-4…
Two charges $2 \mathrm{C}$ and $6 \mathrm{C}$ are separated by a finite distance. If a charge of $-4 \mathrm{C}$ is added to each of them, the initial force of $12 \times 10^3 \mathrm{~N}$ will change to
$4 \times 10^3 \mathrm{~N}$ (repulsion)
$4 \times 10^2 \mathrm{~N}$ (repulsion)
$6 \times 10^3 \mathrm{~N}$ (attraction)
$4 \times 10^3 \mathrm{~N}$ (attraction)
Solution
$\begin{aligned}
& \text { Initial force, } F=k \frac{q_1 q_2}{r^2} \\
& F_1=k \cdot \frac{(2)(6)}{r^2} \\
& \text { New force, } F_2=k \frac{(2-4)(6-4)}{r^2}=k \frac{(-2)(2)}{r^2} \\
& \therefore \quad \frac{F_1}{F_2}=\frac{(2 \times 6)}{(-2 \times 2)} \\
& \frac{12 \times 10^3}{F_2}=-3 \\
& F_2=-4 \times 10^3 \mathrm{~N}
\end{aligned}$
Force is negative, so it will be of attraction.