Question

A uniformly charged ring charge +q has radius R. A uniformly charged solid sphere (over its total volume) has charge Q is placed on the axis of ring at a distance x = 3 R from centre of ring as shown. Radius of sphere is also R, then – inc in tension of ring is

Answer 1

$\frac{1}{4\pi\epsilon_0} \frac{Qq}{20\pi\sqrt{10}R^2}$

Answer 2

To determine the increase in tension of the ring, we need to calculate the radial force exerted by the charged sphere on the ring.

1. Identify the forces: The uniformly charged solid sphere (charge Q, radius R) is placed on the axis of the ring (charge +q, radius R) at a distance x = 3R from the center of the ring. Since the ring is at a distance x = 3R from the center of the sphere, which is greater than the sphere's radius R, the sphere can be treated as a point charge Q located at its center for calculating the electric field at the ring.

2. Calculate the force on an infinitesimal charge element of the ring: Consider an infinitesimal charge element $dq$ on the ring. The distance from the center of the sphere (at x=3R) to any point on the ring (at radius R in the y-z plane) is: $r' = \sqrt{x^2 + R^2} = \sqrt{(3R)^2 + R^2} = \sqrt{9R^2 + R^2} = \sqrt{10R^2} = \sqrt{10}R$.

   The magnitude of the force $dF$ exerted by the sphere on the charge element $dq$ is given by Coulomb's Law: $dF = \frac{1}{4\pi\epsilon_0} \frac{Q dq}{r'^2} = \frac{1}{4\pi\epsilon_0} \frac{Q dq}{(\sqrt{10}R)^2} = \frac{1}{4\pi\epsilon_0} \frac{Q dq}{10R^2}$.

3. Resolve the force into components: This force $dF$ acts along the line connecting the center of the sphere to the charge element $dq$. We need the component of this force that acts radially outwards from the center of the ring, as this component contributes to the tension. Let $\alpha$ be the angle between the line connecting the sphere's center to $dq$ and the axis of the ring (x-axis). From the geometry, $\sin\alpha = \frac{R}{r'} = \frac{R}{\sqrt{10}R} = \frac{1}{\sqrt{10}}$. The radial component of the force $dF_r$ is $dF \sin\alpha$: $dF_r = dF \sin\alpha = \left(\frac{1}{4\pi\epsilon_0} \frac{Q dq}{10R^2}\right) \left(\frac{1}{\sqrt{10}}\right)$ $dF_r = \frac{1}{4\pi\epsilon_0} \frac{Q dq}{10\sqrt{10}R^2}$.

   This radial force component is uniform for all charge elements around the ring.

4. Calculate the radial force per unit length: The total charge on the ring is $q$, and its circumference is $2\pi R$. So, the linear charge density of the ring is $\lambda = \frac{q}{2\pi R}$. An infinitesimal length element of the ring is $dl$. The charge on this element is $dq = \lambda dl$. The radial force per unit length, $f_r$, is $\frac{dF_r}{dl}$: $f_r = \frac{1}{dl} \left(\frac{1}{4\pi\epsilon_0} \frac{Q (\lambda dl)}{10\sqrt{10}R^2}\right) = \frac{1}{4\pi\epsilon_0} \frac{Q \lambda}{10\sqrt{10}R^2}$ Substitute $\lambda = \frac{q}{2\pi R}$: $f_r = \frac{1}{4\pi\epsilon_0} \frac{Q (q/2\pi R)}{10\sqrt{10}R^2} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{20\pi\sqrt{10}R^3}$.

5. Calculate the tension in the ring: For a ring subjected to a uniform outward radial force per unit length $f_r$, the tension $T$ in the ring is given by $T = f_r R$. $T = \left(\frac{1}{4\pi\epsilon_0} \frac{Qq}{20\pi\sqrt{10}R^3}\right) R$ $T = \frac{1}{4\pi\epsilon_0} \frac{Qq}{20\pi\sqrt{10}R^2}$.

   This tension is the increase in tension due to the presence of the sphere.