JEE Main 2023 — Electrostatics Question with Solution

The repulsive force between the two charges is given by Coulomb's law:

$$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2},$$

where $F$ is the force, $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and ${ϵ}_0$ is the electric constant.

To maximize the force, we need to maximize the product $q_1{q}_2$. Let $q_1$ be the charge on one part and $q_2$ be the charge on the other part. Then we have $q_1+q_2=10\mu \mathrm{C}$, since the total charge is $10\mu \mathrm{C}$.

The distance between the two charges is $r=1\mathrm{cm}=0.01\mathrm{m}$. To maximize the force, we need to maximize $q_1{q}_2$, subject to the constraint that $q_1+q_2=10\mu \mathrm{C}$.

We can use the method of Lagrange multipliers to find the values of $q_1$ and $q_2$ that maximize $q_1{q}_2$ subject to the constraint $q_1+q_2=10\mu \mathrm{C}$. The Lagrangian is given by $$\mathcal{L}=q_1{q}_2-\lambda (q_1+q_2-10\mu \mathrm{C}),$$ where $\lambda$ is the Lagrange multiplier.

Taking the partial derivatives of $\mathcal{L}$ with respect to $q_1$, $q_2$, and $\lambda$, and setting them equal to zero, we get:

$$q_2-\lambda =0$$

$$q_1-\lambda =0$$

$$q_1+q_2=10\mu \mathrm{C}$$

Solving for $q_1$ and $q_2$, we get $q_1=q_2=5\mu \mathrm{C}$.

Therefore, the charges of the two parts are both $5\mu \mathrm{C}$.

Therefore, to maximize the repulsive force between the two charges, we need to divide the $10\mu \mathrm{C}$ charge into two equal parts of $5\mu \mathrm{C}$ each, and place them $1\mathrm{cm}$ apart.