Question

Two-point charges $+3\mu \text{C}$ and $+8\mu \text{C}$ repel each other with a force of $40\text{N}$. If a charge of $-5\mu \text{C}$ is added to each of them, then the force between them will become

• A. $-10\text{N}$
• B. $+10\text{N}$
• C. $+20\text{N}$
• D. $-20\text{N}$

Answer 1

The correct option is A $-10\text{N}$

Given:
Charge $q_1=3\mu \text{C}$

Charge $q_2=8\mu \text{C}$

Now, by the Coulomb’s law,

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

In first case,

$40=\frac{1}{4\pi {\epsilon }_0}\frac{3\times 8}{r^2}......\left(1\right)$

When the third charge $q_3=-5\mu \text{C}$ is added to each, then the new charges on $q_1$ and $q_2$ will be :

$q_1^{{}^{\prime }}=3-5=-2\mu \text{C}$

$q_2^{{}^{\prime }}=8-5=3\mu \text{C}$

In second case $F^{{}^{\prime }}=\frac{1}{4\pi {\epsilon }_0}\frac{(-2\times 3)}{r^2}.....\left(1\right)$

From equation $\left(1\right)$ & $\left(2\right)$,

$\frac{F^{{}^{\prime }}}{40}=\frac{(-2\times 3)}{3\times 8}$

$F^{{}^{\prime }}=-10\text{N}$