Question

Two identical metallic spheres, having unequal opposite charges are placed at a distance of $0.50m$ apart in air. After bringing them in contact with each other, they are again placed at the same distance apart. Now the force of repulsion between them is $0.108N$. Calculate the final charge on each of them in horizontal.

Answer 1

When the two charges are brought in contact with each other, charge redistribution takes place and both the spheres acquire equal charge of the same polarity. This is the reason why a force of repulsion acts between them. We shall use the formula of force between these two charges to find the charge on both of them.

Complete answer:
Let the initial charges on the spheres be ${{q}_{1}}$ and ${{q}_{2}}$ respectively.

After bringing them in contact with each other, the spheres acquire equal charges, let us say, $Q$ due to charge redistribution.

The force on any charged body is given as:
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Where,
${{\varepsilon }_{0}}=$ permittivity of free space or vacuum and ${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}F{{m}^{-1}}$
${{q}_{1}}=$ charge on one body
${{q}_{2}}=$ charge on other body
$r=$ distance between the two bodies
Now, $\dfrac{1}{4\pi {{\varepsilon }_{0}}}=k$ and $k$ is a constant which is equal to $9\times {{10}^{9}}$, thus, $\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}$
We have, ${{q}_{1}}={{q}_{2}}=Q$, $r=0.50m$ as the spheres are put back to their initial positions and given that $F=0.108N$.
Substituting all these values, we get:
$0.108=9\times {{10}^{9}}.\dfrac{Q.Q}{{{\left( 0.50 \right)}^{2}}}$
$\begin{aligned} & \Rightarrow 0.108=9\times {{10}^{9}}.\dfrac{{{Q}^{2}}}{{{\left( 0.50 \right)}^{2}}} \\\ & \Rightarrow {{Q}^{2}}=\dfrac{0.108\times 0.25}{9\times {{10}^{9}}} \\\ & \Rightarrow {{Q}^{2}}=\dfrac{27\times {{10}^{-3}}}{9\times {{10}^{9}}} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow {{Q}^{2}}=3\times {{10}^{-12}}C \\\ & \Rightarrow Q=\pm \sqrt{3\times {{10}^{-12}}} \\\ & \Rightarrow Q=\pm \sqrt{3}\times {{10}^{-6}}C \\\ \end{aligned}$
Therefore, the final charge on both the spheres is either $+\sqrt{3}\mu C$ or $-\sqrt{3}\mu C$.

Note:
When a charged body is isolated, it does not experience any force and stays in equilibrium. When another charged body comes in vicinity of the first one, both the charges apply forces on each other. The first charge experiences force from the second one and the second one experiences force from the first one. If the two charges are of the same polarity, then they repel each other but if they have polarity of opposite nature, then they exert attractive forces on each other.