Question

Two small insulated charged spheres of the same size are placed in air so that the distance between their centers is $20cm$.Each sphere carries a charge of $+0.6\mu C$.
$\left(1\right)$ Find the force of repulsion between them.
$\left(2\right)$ If each sphere were to carry double the charge and kept double the distance apart, what would be the force ?
$\left(3\right)$ If in $\left(1\right)$ the spheres were in a dielectric medium of constant $5$, what would be the force?
$\left(4\right)$ An uncharged similar sphere is brought into contact with the first sphere, then with the second sphere and then removed from the site, What would be the force between the spheres now ?

Answer 1

1. Given, $q_1=q_2=0.6\mu C$

Distance,$r=20cm=0.2m$

Force of repulsion, $F=K\frac{q_1{q}_2}{r^2}$

$=9\times {10}^9\frac{0.6\times 0.6\times {10}^{-12}}{{\left(0.2\right)}^2}N$

$=\frac{324}{4}\times {10}^{-3}N$

$=81\times {10}^{-3}N$

When each charges are doubled by magnitude and distance between the two spheres is also doubled $q_1=q_2=1.2\times {10}^{-6}C$ and Distance,$r=2\times 0.2=0.4m$

So force is, $F=\frac{9\times {10}^9\times 1.2\times 1.2\times {10}^{-12}}{{\left(0.4\right)}^2}N$

$=\frac{1296\times {10}^{-3}}{16}N$

$=81\times {10}^{-3}N$

$3$. Given dielectric constant,$k=5$

So the force of repulsion is, $F=\frac{K}{k}\frac{q_1{q}_2}{r^2}$

$F=\frac{9\times {10}^9}{5}\frac{0.6\times 0.6\times {10}^{-12}}{{\left(0.2\right)}^2}N$

$F=\frac{3.24\times {10}^{-3}}{0.2}N$

$F=16.2\times {10}^{-3}N$

$4$. Distance between the spheres, A and B, $r=0.2m$

Initially, the charge on each sphere, $q=0.6\mu C$

When sphere A is touched with an uncharged sphere C, $q/2$ amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is $q/2$.

When sphere C with charge $q/2$ is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as, $1/2(q+q/2)=3q/4$.

Force of repulsion between sphere A having charge $q/2$ and sphere B having $3q/4$ charge is

$F=\frac{K\frac{q}{2}\frac{3q}{4}}{r^2}=\frac{9\times {10}^9\times 3q^2}{8\times 0.04}=\frac{9\times {10}^9\times 3\times {\left(0.6\right)}^2\times {10}^{-12}}{8\times 0.04}=30.37\times {10}^{-3}N$