Question

Two identical metallic spheres, having unequal opposite charges are placed at a distance of $0.90\mathrm{m}$ apart. 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 $0.025\mathrm{N}$. Calculate the final charge on each of them.

Answer 1

Step 1: Given parameters.

Let the initial charges be $q_1$ and $q_2$ respectively.

The force of repulsion between the charges is $0.025\mathrm{N}$.

After they come in contact, the charges are rearranged such that they acquire the same charge.

Consider the charge on each of them is $Q$.

Step 2: Calculate the charge.

They are again brought apart at a distance of $0.90\mathrm{m}$.

Use the coulombs law given as

$F=\frac{{\mathrm{KQ}}^2}{{\mathrm{r}}^2}$

Where, $F$ is the force of repulsion,$Q$ is the charge, and, $\mathrm{K}$ is the Columb's constant.

Assume,

$\mathrm{K}=9\times {10}^9\mathrm{N}‐{\mathrm{m}}^2/{\mathrm{C}}^2$

Step 3: Replace the values in the formula

$\begin{array}{rcl}F& =& \frac{{\mathrm{KQ}}^2}{{\mathrm{r}}^2}\\ & \Rightarrow & \mathbf{}\mathbf{}0.025=\frac{9\times {10}^9\times Q^2}{0.9^2}\\ & \Rightarrow & Q^2=\frac{0.025\times 0.9^2}{9\times {10}^9}\\ & \Rightarrow & Q=1.5\times {10}^{-6}C\end{array}$

Therefore, the final charge on each of them is $1.5\times {10}^{-6}\mathrm{C}$.