JEE Physics 2026 Question with Solution

Given: Point charge $q = 1 \times 10^{-6} \, \text{C}$, wire length $L = 0.10 \, \text{m}$, total charge on wire $Q = 24 \times 10^{-6} \, \text{C}$, nearest end distance $0.02 \, \text{m}$, and

$$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$$

Find: Force between the point charge and the uniformly charged wire.

Step 1: Find linear charge density of the wire.

$$\lambda = \frac{Q}{L} = \frac{24 \times 10^{-6}}{0.10} = 2.4 \times 10^{-4} \, \text{C/m}$$

Step 2: Consider an element of wire. Let an element $dx$ be at a distance $x$ from the point charge.

$$dq = \lambda \, dx$$

Step 3: Write expression for force due to the element.

$$dF = \frac{1}{4\pi\varepsilon_0} \frac{q \, dq}{x^2} = 9 \times 10^9 \frac{q\lambda}{x^2} \, dx$$

Step 4: Set limits of integration. The nearest end is at $x = 0.02 \, \text{m}$ and the far end is at $x = 0.12 \, \text{m}$.

Step 5: Integrate to find total force.

$$F = 9 \times 10^9 \, q\lambda \int_{0.02}^{0.12} \frac{dx}{x^2}$$ $$F = 9 \times 10^9 \times (1 \times 10^{-6}) \times (2.4 \times 10^{-4}) \left[ -\frac{1}{x} \right]_{0.02}^{0.12}$$ $$F = 9 \times 10^9 \times 2.4 \times 10^{-10} \left( \frac{1}{0.02} - \frac{1}{0.12} \right)$$ $$F = 2.16 \times (50 - 8.33) \approx 90 \, \text{N}$$

Therefore, the force between the point charge and the wire is $90 \, \text{N}$.

Given: The wire carries uniform charge, so the appropriate method is to divide it into small charge elements and add their Coulomb forces.

Because the wire is collinear with the point charge, all elemental forces act along the same line, so their magnitudes can be integrated directly.

For an element at distance $x$,

$$dq = \lambda \, dx$$

and

$$dF = k \frac{q \, dq}{x^2}$$

where

$$k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$$

Substituting $dq = \lambda dx$ gives

$$dF = 9 \times 10^9 \frac{q\lambda}{x^2} \, dx$$

The wire begins $0.02 \, \text{m}$ from the charge and ends at

$$0.02 + 0.10 = 0.12 \, \text{m}$$

So,

$$F = 9 \times 10^9 q\lambda \int_{0.02}^{0.12} x^{-2} \, dx$$

Using

$$\int x^{-2} \, dx = -\frac{1}{x}$$

we obtain the same final value,

$$F \approx 90 \, \text{N}$$

Hence, the numerical answer is 90.