JEE Physics 2023 Question with Solution

Given: A charge $10 \, \mu C$ is at the origin and the net electric field at point $P$ on the x-axis at $x = 2 \, \text{cm}$ is zero. Another charge $40 \, \mu C$ is to be placed at $x_0$.

Find: The location $x_0$ of the $40 \, \mu C$ charge.

For the net electric field at $P$ to be zero, the electric fields due to the two positive charges must be equal in magnitude and opposite in direction.

So,

From the figure and field direction requirement, the $40 \, \mu C$ charge must be to the right of $P$, hence

Therefore, the correct option is A and the required location is $x = 6 \, \text{cm}$.

At $x = 2 \, \text{cm}$, the field due to the $10 \, \mu C$ charge at the origin points toward the positive x-direction because the source charge is positive.

To cancel this field, the field due to the $40 \, \mu C$ charge at $P$ must point toward the negative x-direction. That happens only if the $40 \, \mu C$ charge is placed to the right of $P$.

Although solving

gives $x_0 = 6 \, \text{cm}$ or $x_0 = -2 \, \text{cm}$, only $x_0 = 6 \, \text{cm}$ satisfies the required field direction. Also, the solution's marks option D while writing $x = 6 \, \text{cm}$; this is a label mismatch in the source, and the correct mapped option here is A.