JEE Physics 2023 Question with Solution

Given: Total charge is $q = 10 \, \mu \text{C}$ and the two parts are kept at a fixed distance $r = 1 \, \text{cm}$.

Find: The division of charge for which the repulsive force is maximum.

Using Coulomb's law,

$$F = \frac{k \, x \, (q-x)}{r^2}$$

where one part is $x$ and the other is $(q-x)$.

To maximize force, differentiate with respect to $x$ and set it equal to zero:

$$\frac{dF}{dx} = \frac{d}{dx}\left(\frac{k \, x \, (q-x)}{r^2}\right) = \frac{k}{r^2}(q-2x) = 0$$

So,

$$q - 2x = 0$$

Hence,

$$x = \frac{q}{2} = \frac{10 \, \mu \text{C}}{2} = 5 \, \mu \text{C}$$

The other part is,

$$q - x = 10 \, \mu \text{C} - 5 \, \mu \text{C} = 5 \, \mu \text{C}$$

Therefore, the charges of the two parts are $5 \, \mu \text{C}$ and $5 \, \mu \text{C}$. The correct option is D. The solution mentions option C, but the worked solution clearly gives $5 \, \mu \text{C}$ and $5 \, \mu \text{C}$, which matches option D.

Given: Total charge is fixed and distance is fixed.

Find: When the product of the two charge parts is maximum.

Since

$$F \propto x(q-x)$$

for fixed $q$ and fixed $r$, the force is maximum when the product $x(q-x)$ is maximum. For a fixed sum, the product is maximum when the two numbers are equal.

Thus,

$$x = q-x = \frac{q}{2} = 5 \, \mu \text{C}$$

Therefore, the two charges are $5 \, \mu \text{C}$ and $5 \, \mu \text{C}$. The correct option is D.