JEE Physics 2025 Question with Solution

Given: $q_1$ and $q_2$ are separated by $30 \, \text{cm}$. Charge $q_3$ moves from C to D on a circular path of radius $40 \, \text{cm}$ centered at A, where $q_1$ is placed.

Find: The value of $K$ if the change in potential energy is $\frac{q_3 K}{4 \pi \epsilon_0}$.

At C and D, the distance of $q_3$ from $q_1$ remains the same, equal to $40 \, \text{cm}$. Therefore, the potential energy contribution due to interaction of $q_3$ with $q_1$ does not change.

From the geometry shown:

• Distance of C from $q_2$ is

• Distance of D from $q_2$ is

Hence,

and

where $k = \frac{1}{4 \pi \epsilon_0}$.

Therefore, the change in potential energy is

As written in the provided solution, this is expressed in the form

and the final result obtained there is

Therefore, the correct option is A.

Given: $AC = 40 \, \text{cm}$, $AB = 30 \, \text{cm}$, and C to D is a circular arc centered at A.

Find: The coefficient $K$ in $\Delta U = \frac{q_3 K}{4 \pi \epsilon_0}$.

Because A is the center of the circular path, both C and D are at the same distance from A. So the term involving $q_1$ cancels in the difference $U(D) - U(C)$.

Now use the distances from $q_2$:

Also, since $AD = 40 \, \text{cm}$ and $AB = 30 \, \text{cm}$,

So only the interaction of $q_3$ with $q_2$ changes:

The extracted solution concludes that this corresponds to

Therefore, the correct option is A.