JEE Main 2024 — 27 January, Shift 1 — Question 49

The tension in the ring is primarily caused by the electrostatic repulsion from the point charge located at its center. While a uniformly charged ring also experiences self-repulsion, in the context of such problems, the question often refers to the tension induced by the external force (in this case, from the central point charge $Q_2$ on the distributed charge $Q_1$).

Given values: Radius of the ring, $R = 30 \text{ cm} = 0.3 \text{ m}$. Charge uniformly distributed over the ring, $Q_1 = 2\pi \text{ C}$. Point charge at the center of the ring, $Q_2 = 30 \text{ pC} = 30 \times 10^{-12} \text{ C}$. Coulomb's constant, $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$.

First, calculate the linear charge density $\lambda$ of the ring:

$$\lambda = \frac{Q_1}{2\pi R}$$

Substitute the given values for $Q_1$ and $R$:

$$\lambda = \frac{2\pi \text{ C}}{2\pi \times 0.3 \text{ m}} = \frac{1}{0.3} \text{ C/m} = \frac{10}{3} \text{ C/m}$$

Now, consider a small element of the ring with charge $dq_1 = \lambda R d\theta$. This element experiences a repulsive force from the central charge $Q_2$. The magnitude of this force $dF$ is given by Coulomb's law:

$$dF = \frac{k Q_2 dq_1}{R^2}$$

Substitute $dq_1 = \lambda R d\theta$:

$$dF = \frac{k Q_2 (\lambda R d\theta)}{R^2} = \frac{k Q_2 \lambda d\theta}{R}$$

This force $dF$ acts radially outwards on the small element.

For the ring to be in equilibrium (not getting deformed), the radially outward force on this small element must be balanced by the radial component of the tension ($T$) in the ring. If $T$ is the tension in the ring, then the inward radial component of the tension from the two ends of the small segment of angle $d\theta$ is $2T \sin(d\theta/2)$. For small angles, $\sin(d\theta/2) \approx d\theta/2$. So, the inward radial component of tension is $2T \left(\frac{d\theta}{2}\right) = T d\theta$.

Equating the outward electrostatic force and the inward radial component of tension for equilibrium:

$$T d\theta = \frac{k Q_2 \lambda d\theta}{R}$$

Divide by $d\theta$:

$$T = \frac{k Q_2 \lambda}{R}$$

Now, substitute the numerical values for $k$, $Q_2$, $\lambda$, and $R$:

$$T = \frac{(9 \times 10^9 \text{ Nm}^2/\text{C}^2) \times (30 \times 10^{-12} \text{ C}) \times (\frac{10}{3} \text{ C/m})}{0.3 \text{ m}}$$ $$T = \frac{(9 \times 10^9) \times (10 \times 10^{-12}) \times 10}{0.3}$$ $$T = \frac{9 \times 10^9 \times 10^{-11}}{0.3} \times 10$$ $$T = \frac{9 \times 10^{-2}}{0.3} \times 10$$ $$T = \frac{0.09}{0.3} \times 10$$ $$T = 0.3 \times 10$$ $$T = 3 \text{ N}$$

The tension in the ring is $3 \text{ N}$.