Question

A ring of radius $R$ having a linear charge density $\lambda$ moves towards a solid imaginary sphere of radius $\frac{R}{2}$, so that the centre of the ring passes through the centre of sphere. The axis of the ring is perpendicular to the line joining the centres of the ring and the sphere. The maximum flux through the sphere in this process is

• A. $\frac{\lambda \pi R}{3{\epsilon }_0}$
• B. $\frac{\lambda R}{2{\epsilon }_0}$
• C. $\frac{\lambda R}{{\epsilon }_0}$
• D. $\frac{\lambda \pi R}{4{\epsilon }_0}$

Answer 1

The correct option is A $\frac{\lambda \pi R}{3{\epsilon }_0}$

Electric flux will be maximum when maximum length of ring is inside the sphere.
This will occur when the length of the chord $\text{AB}$ is maximum.

Now, maximum length of chord $\text{AB}$ can be the diameter of the sphere.

Since, diameter of sphere is equal to radius of ring, hence it will form an equilateral triangle, each angle having ${60}^{\circ }$.

In this case length of the arc of ring inside the sphere can be calculated as,

Arc length $=\frac{2\pi R}{{360}^{\circ }}\times {60}^{\circ }=\frac{\pi }{3}R$

Thus, charge $\left(q\right)$ on the arc in this situation is given as,
$q=\text{Arc length}\times \lambda$

$\Rightarrow q=\frac{R\pi }{3}\lambda$

Thus, the maximum flux ${\varphi }_{max}$ through solid sphere is
${\varphi }_{max}=\frac{q}{{\epsilon }_0}$

$\Rightarrow {\varphi }_{max}=\frac{\frac{R\pi }{3}\lambda }{{\epsilon }_0}$

$\Rightarrow {\varphi }_{max}=\frac{R\pi \lambda }{3{\epsilon }_0}$

Hence, option (d) is correct answer.