Question

A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring with one of its ends coinciding with the centre of ring. The total charge of the ring is equal to Q.The charge of the thread (per unit length) is equal to $\lambda$. Find the force of interaction between the ring and the thread.

• A. Zero
• B. $\frac{\lambda Q}{2\pi {ϵ}_{0}R}$
• C. $\frac{\lambda Q}{4\pi {ϵ}_{0}R}$
• D. $\frac{\lambda Q}{\pi {ϵ}_{0}R}$

Answer 1

The correct option is C $\frac{\lambda Q}{4\pi {ϵ}_{0}R}$

The electric field strength due to ring at a point on its axis (say x-axis) at distance x from the centre of the ring is given by
$E\left(x\right)=\frac{Qx}{4\pi {ϵ}_0(R^2+x^2{)}^{3/2}}$
And from symmetry $\overrightarrow{E}$ at every point on the axis is directed along the x-axis Let us consider an element dx on thread which carries the charge $dq=\lambda dx$. The electric force experienced by the element in the field of ring is
$dF=\left(\lambda dx\right)E\left(x\right)=\frac{\lambda Qxdx}{4\pi {ϵ}_0(R^2+x^2{)}^{3/2}}$
On integrating we get, $F=\frac{\lambda Q}{4\pi {ϵ}_{0}R}$