Question

Two identical conducting spheres of a small radius are charged and placed 25 cm apart. It is observed that they attract each other with a force 1.728 N. The spheres are now connected by a thin conducting wire and the wire is then removed. It is observed that the spheres now repel each other with a force 0.576 N. Find the initial charges on the sphere.

Answer 1

Since the spheres attract initially, they carry charges of opposite nature, say $q_1$ and $q_2.$ Magnitude of force, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
Here,
$r=25cm=25\times {10}^{-2}m$
F=1.725 N
$\therefore 1.728=9\times {10}^9\frac{q_1{q}_2}{(25\times {10}^{-2}{)}^2}$
or $q_1{q}_2=12\times {10}^{-12}$ .....(1)
When the spheres are connected by a conducting connecting wire, charges on them redistribute so as to attain a common electric potential. If $q_1^{{}^{\prime }}and-q_2^{{}^{\prime }}$ are the charges after redistribution, equality of electric potential requires
$\frac{1}{4\pi {\epsilon }_0}\frac{q_1^{{}^{\prime }}}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1^{{}^{\prime }}}{R}$
$[PotentialonsurfaceV=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^{{}^{\prime }}},whereRistheradius]$
Since there is no loss of charge during redistribution, the total charge on the system remains the same.
Here,
$q_1^{{}^{\prime }}+q_2^{{}^{\prime }}=q_1+(-q_2)=q_1-q_1$
but $q_1^{{}^{\prime }}=q_2^{{}^{\prime }}$
$\therefore q_1^{{}^{\prime }}=q_2^{{}^{\prime }}=\frac{q_1-q_2}{2}$
The charges are now observed to repel with a force 0.576 N.
$\therefore \frac{1}{4\pi {\epsilon }_0}\frac{q_1^{{}^{\prime }}+q_2^{{}^{\prime }}}{r^2}=0.576$
or
$9\times {10}^9\frac{\left(\frac{q_1-q_2}{2}\right)}{(25\times {10}^{-2}{)}^2}=0.576$
$\therefore q_1-q_2=4\times {10}^{-6}$ ....(2)
Substituting $q_2$ from Eq. (2) in Eq. (1)
$q_1(q_1-4\times {10}^{-6})=12\times {10}^{-12}$
$q_1^2-(4\times {10}^{-6})q_1-12\times {10}^{-12}=0$
Solving, the roots are $6\times {10}^{-6}Cand-2\times {10}^{-6}C$
Hence, the two charges are $6\times {10}^{-6}Cand-2\times {10}^{-6}C.$