Question

A point charge $Q$ is located at centre of a fixed thin ring of radius $R$ with uniformly distributed charge $-Q$. The magnitude of the electric field strength at the point lying on the axis of the ring at a distance $x$ from the centre is $(x>>R)$

• A. $\frac{Q{R}^2}{8\pi {\epsilon }_0{x}^4}$
• B. $\frac{3Q{R}^2}{8\pi {\epsilon }_0{x}^4}$
• C. $\frac{5Q{R}^2}{8\pi {\epsilon }_0{x}^4}$
• D. $\frac{7Q{R}^2}{8\pi {\epsilon }_0{x}^4}$

Answer 1

The correct option is B $\frac{3Q{R}^2}{8\pi {\epsilon }_0{x}^4}$

Electric field at point P due to +Q is

${\in }_1\frac{KQ}{x^2}\to \left(i\right)$

Electric point at point P due to ring is

${\in }_2=\frac{-KQx}{\left(x^2+R^2\right)}\frac{3}{2}\to \left(ii\right)$

Net electric field at P is

$\begin{array}{c}\in ={\in }_1+{\in }_2\\ \therefore \frac{KQ}{x^2}+\frac{-KQx}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\Rightarrow \in =\frac{KQ}{x^2}+\frac{-KQx}{x^3{\left(1+\frac{R^2}{x^2}\right)}^{\frac{3}{2}}}\\ \Rightarrow \in =\frac{KQ}{x^2}\left[1-{\left(1+\frac{R^2}{x^2}\right)}^{\frac{-3}{2}}\right]\to \left(iii\right)\\ \sin ex\gg R,\frac{R^2}{x^2}\ll 1,usingbinomialsweget\\ \in =\frac{KQ}{x^2}\left[12-\left\{1+\left(\frac{-3}{2}\right)\left(\frac{R^2}{x^2}\right)\right\}\right]e\\ \in =\frac{KQ}{x^2}\left[1-1+\frac{3R^2}{2x^2}\right]\\ \in =\frac{3Q{R}^2}{8\pi \in \cdot x^4}\left(K=\frac{1}{4\pi \in o}\right)\end{array}$