Two charged identical metal spheres A and B repel each other with a force of  $3\times {10}^{-5}\mathbf{ }N$. Another identical uncharged sphere C is touched with sphere A and then it is placed midway between A and B. Then the magnitude of Net force on C is

Given data: Two identical metal spheres A and B that are charged repel one another with a force of $2\times {10}^{-5}$ 

Concept used: Application of gauss law.

Detailed solution:

Given that two metal spheres with the same charge, A and B, strongly repel one another, $F=3\times {10}^{-5} \mathrm{N}$.

Then, sphere A contacts sphere c, which is identical, and is positioned halfway between A and B.

Coulomb’s force,  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{R^2}$

This is the interaction of spheres A and B.

q is the charge on spheres A and B, and R is the separation between two charges.

Now that C has touched A, it will receive a charge q/2 , while A keeps its charge of q/2. following that, when it is positioned halfway between A and B force felt by C is:

 $F_r=\text{ force on C due to }\mathrm{B}-\text{ Force on C due to }\mathrm{A}$ 

 $\frac{{\left(\frac{q^2}{2}\right)}^2}{4\pi {\epsilon }_0{\left({\displaystyle \frac{R}{2}}\right)}^2}-\frac{{\displaystyle \frac{q^2}{2}}}{4\pi {\epsilon }_0{\left({\displaystyle \frac{R}{2}}\right)}^2}=-\frac{q^2}{4\pi {\epsilon }_0{R}^2}$  

 $F_c=F=3\times {10}^{-5}N$