JEE Physics 2026 Question with Solution

Given: Source charge $Q = 10^{-8} \, \text{C}$, test charge $q = 2 \times 10^{-6} \, \text{C}$, $k = 9 \times 10^9$, point $A = (4,4,2)$ and point $B = (2,2,1)$.

Find: The work done in moving the charge from $A$ to $B$.

Since the electric field is conservative, the work depends only on the initial and final positions. Use electric potential:

$$V = k\frac{Q}{r}$$

and

$$W = q(V_B - V_A)$$

First calculate the distances from the origin:

$$r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{36} = 6 \, \text{m}$$ $$r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3 \, \text{m}$$

Now the potentials at the two points are:

$$V_A = k\frac{Q}{r_A} = (9 \times 10^9)\frac{10^{-8}}{6} = \frac{90}{6} = 15 \, \text{V}$$ $$V_B = k\frac{Q}{r_B} = (9 \times 10^9)\frac{10^{-8}}{3} = \frac{90}{3} = 30 \, \text{V}$$

Therefore,

$$W_{A \to B} = q(V_B - V_A)$$ $$W_{A \to B} = (2 \times 10^{-6})(30 - 15)$$ $$W_{A \to B} = 30 \times 10^{-6} \, \text{J}$$

Therefore, the work done in moving the charge from point $A$ to point $B$ is $30 \times 10^{-6} \, \text{J}$. The correct option is C.

Given: The charge is moved in the field of a point charge at the origin.

Find: The required work from $A$ to $B$.

The hint emphasizes the sign convention. Work done by the electric field is

$$q(V_A - V_B)$$

whereas work done by an external agent in moving the charge slowly is

$$q(V_B - V_A)$$

Here the solution interprets the question as the latter.

Since point $B$ is closer to the positive source charge than point $A$, its potential must be greater. Thus the required work is positive. After computing

$$V_A = 15 \, \text{V}, \qquad V_B = 30 \, \text{V}$$

we get

$$W = q(V_B - V_A) = 2 \times 10^{-6} \times 15 = 30 \times 10^{-6} \, \text{J}$$

Hence the correct option is C.