Question

A charge q is distributed over a thin ring of radius a. The work of the field forces done in the displacement of a point charge Q from the centre of the ring to infinity will be :

• A. $\text{z}ero$
• B. $\frac{qQ}{4\pi {\epsilon }_{0}a}$
• C. $\text{i}nfinite$
• D. $\frac{qQ}{4\pi {\epsilon }_0{a}^2}$

Answer 1

The correct option is B $\frac{qQ}{4\pi {\epsilon }_{0}a}$

Electric potential at the centre of charged thin ring $=\frac{q}{4\pi {ϵ}_{0}a}$
Hence, potential energy of a particle $Q$ at the centre of this ring $=\frac{Q\cdot q}{4\pi {ϵ}_{0}a}$
Potential energy of particle $Q$ at infinity $=0$
Work done against the field forces to carry this particle to infinity $=\Delta P.E.=\frac{Q\cdot q}{4\pi {ϵ}_{0}a}-0$ $=\frac{Qq}{4\pi {ϵ}_{0}a}$