Question

Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times {10}^{–7}$ C? The radii of A and B are negligible compared to the distance of separation.

What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer 1

Let us consider, the charge on sphere A is $q_A$ and on sphere B is $q_B$. According to the question,
$q_A=6.5\times {10}^{-7}C{q}_B=6.5\times {10}^{-7}C$
r=distance between A and B=50cm= $50\times {10}^{-2}m$
From the Coulomb's law, the force between the two spheres is
$F=\frac{1}{4\pi {\epsilon }_0}.\frac{q_A{q}_B}{r^2}=\frac{9\times {10}^9\times 6.5\times {10}^{-7}\times 6.5\times {10}^{-7}}{(50\times {10}^{-2}{)}^2}=\frac{9\times 6.5\times 6.5\times {10}^{-5}}{50\times 50\times {10}^{-4}}=1.521\times {10}^{-2}N$
Thus, the force between A and B is $1.521\times {10}^{-2}N,$ this force is repulsive in nature because the charges are similar (positive) in nature.

According to the question, if the charge is doubled
$q_A^{\prime }=2q_{A}and{q}_B^{\prime }=2q_B$
Distance between them is halved i.e., $r^{\prime }=\frac{r}{2}$
Now, the force between the two spheres is
$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}.\frac{q_A^{\prime }{q}_B^{\prime }}{r^{\prime }}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(2q_A\right)\left(2q_B\right)}{(r/2{)}^2}=\frac{1}{4\pi {\epsilon }_0}.\frac{4q_A{q}_B}{r^2/4}=16\frac{1}{4\pi {\epsilon }_0}\frac{q_A{q}_B}{r^2}=16F=16\times 1.521\times {10}^{-2}=0.24N$
This force is also repulsive in nature because both the charges are similar (positive) in nature.