Question

Two identical conducting spheres with negligible volume have $2.1nC$ and $-0.1nC$ charges, respectively. They are brought into contact and then separated by a distance of $0.5m$. The electrostatic force acting between the spheres is _______$\times {10}^{-9}N$. [Given : $4{\mathrm{\pi \epsilon }}_0=\frac{1}{9}\times {10}^{-9}$ SI unit]

Answer 1

Step 1: Given data

Initial charges in the spheres are $2.1nC$ and $-0.1nC$

Later, they are bought into contact and then and then separated by a distance of $0.5m$

Step 2: To find

The electrostatic force between the spheres

Step 3: Formula and calculation

Initially, the charges are arranged as in the first figure.

Then they are bought into contact and then separated by a distance $0.5m$. Then the charge distribution will be as shown in the second figure.

When the charge bought in contact with each other, they balance the charge on each other such that the charge on each body after they separated is equal to half of the net charge of the system.

The electrostatic force acting between the spheres is given by

$F_e=k\frac{q_1{q}_2}{r^2}$

Where $k$ is constant, $q_1$ and $q_2$ are the charges on the sphere finally and $r$ is the distance between them.

$\therefore F_e=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r^2}\left(\because k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)$

We know,

$$\begin{gathered} 4{\mathrm{\pi \epsilon }}_0=\frac{1}{9}\times {10}^{-9}SIunits \\ {q}_1=q_2=1\times {10}^{-9}C \\ r=0.5m \\ \Rightarrow F_e=9\times {10}^9\times \frac{1\times {10}^{-9}\times 1\times {10}^{-9}}{0.5^2} \\ \Rightarrow F_e=36\times {10}^{-9}N \end{gathered}$$

Thus, the electrostatic force acting between the spheres is $36\times {10}^{-9}N$.