Question

A thin cylindrical wire of radius $r$ uniformly charged with charge $Q$ and fold into circle of radius $R(>>r)$. Now a point charge $q$ is placed at the centre of this circle. Find increment in stress of the wire after placing charge $q$. $\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)$

Answer 1

$\frac{qQ}{8\pi^3 \varepsilon_0 R^2 r^2}$

Answer 2

The problem asks for the increment in stress of a thin cylindrical wire, uniformly charged with charge $Q$ and folded into a circle of radius $R(>>r)$, after a point charge $q$ is placed at its center.

1. Linear Charge Density ($\lambda$) of the wire: The total charge on the wire is $Q$, and it is distributed uniformly over the circumference of the circle, which is $2\pi R$. Therefore, the linear charge density is: $\lambda = \frac{Q}{2\pi R}$

2. Electrostatic Force on a small element of the ring: Consider a small element of the wire subtending an angle $d\theta$ at the center. The length of this element is $dl = R d\theta$. The charge on this small element is $dQ = \lambda dl = \lambda R d\theta$. The electrostatic force $dF$ exerted by the central charge $q$ on this element $dQ$ is directed radially outwards. Using Coulomb's law: $dF = k \frac{q dQ}{R^2} = k \frac{q (\lambda R d\theta)}{R^2} = k \frac{q \lambda d\theta}{R}$ where $k = \frac{1}{4\pi \varepsilon_0}$.

3. Increment in Tension ($T$) in the wire: Let $T$ be the tension developed in the wire due to the outward electrostatic force. Consider the equilibrium of the small element of the ring. The tension forces $T$ act tangentially at the ends of the element. The resultant inward radial component of these tension forces balances the outward electrostatic force $dF$. For a small angle $d\theta$, the inward radial force due to tension is $2T \sin\left(\frac{d\theta}{2}\right) \approx 2T \left(\frac{d\theta}{2}\right) = T d\theta$. Equating the inward tension force component with the outward electrostatic force: $T d\theta = dF$ $T d\theta = k \frac{q \lambda d\theta}{R}$ $T = k \frac{q \lambda}{R}$ Now, substitute the expression for $\lambda$: $T = k \frac{q}{R} \left(\frac{Q}{2\pi R}\right) = k \frac{qQ}{2\pi R^2}$ Substitute $k = \frac{1}{4\pi \varepsilon_0}$: $T = \frac{1}{4\pi \varepsilon_0} \frac{qQ}{2\pi R^2} = \frac{qQ}{8\pi^2 \varepsilon_0 R^2}$ This is the increment in tension in the wire.

4. Increment in Stress ($\sigma$) of the wire: Stress is defined as force per unit cross-sectional area. The force here is the tension $T$. The wire is cylindrical with radius $r$. So, its cross-sectional area $A$ is: $A = \pi r^2$ The increment in stress $\sigma$ is: $\sigma = \frac{T}{A}$ $\sigma = \frac{\frac{qQ}{8\pi^2 \varepsilon_0 R^2}}{\pi r^2}$ $\sigma = \frac{qQ}{8\pi^3 \varepsilon_0 R^2 r^2}$

The condition $R(>>r)$ ensures that the wire can be treated as a line charge for calculating electrostatic forces and that the tension is uniform across the wire's cross-section.

The final answer is $\boxed{\frac{qQ}{8\pi^3 \varepsilon_0 R^2 r^2}}$.