Question

Two point charges of +2 mu C and +6 muC repel each other with a force of 12N. If each is given an additional charge of -4 mu C, what will be the new force?

Answer 1

Answer:

-4 N

Explanation:

adding -4 to each charge then they becomes to -2mu C and 2muC so then force acts b/w then becomes to -4N

Answer 2

Answer:

New force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is$4 \times 10^{-6} \mathrm{~N} .$

Explanation:

Given: Two point charges of +2 mu C and +6 mu C repel each other with a force of 12N.

To find: New force

Solution:

The old charges are $+2 \mu \mathrm{C}$ and $+\mathbf{6} \mu \mathrm{C}$. If an additional charge $-4 \mu \mathrm{C}$ is added to that then the new charges will be $-2 \mu C , 2 \mu C$ respectively.

The force between the two old point charges is $12 \mathrm{~N}$. As they both are positive charges, they repel each other. So the distance between the two charges can be found in coulomb's law.

$F=k \frac{q q^{\prime}}{r^{2}}$

$\begin{aligned}&12=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times 6 \times 10^{-6}}{r^{2}} \\&r^{2}=\frac{9 \times 2 \times 6 \times 10^{3}}{12}=9 \times 10^{3}\end{aligned}$

The distance found can be used in the force between the new charges $-2 \mu \mathbf{C} , 2 \mu \mathrm{C} .$

$\begin{aligned}&F=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times-2 \times 10^{-6}}{9 \times 10^{3}} \\&F=-\frac{9 \times 2 \times 2 \times 10^{9-6-6-3}}{9}=-4 \times 10^{-6} \mathrm{~N}\end{aligned}$

So, the new force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is$4 \times 10^{-6} \mathrm{~N} .$