Question

25) A non - conducting ring of radius 0.5 m carries a total charge of $1.11\times {10}^{-10}C$ distributed non - uniformly on its circumference producing an electric field $\overrightarrow{E}$ everywhere in space. The value of the line integral ${\int }_{t=\infty }^{t=0}-\overrightarrow{E}.d\overrightarrow{\mathcal{l}}(\mathcal{l}=0,$ being the centre of the ring ) in volt is.
(A) + 2
(B) - 1
(C) - 2
(D) 0

Answer 1

Dear Student,
$$\begin{gathered} weareassumingheresimplestcaseataxisofthering \\ soE=\frac{Q}{4\pi {\epsilon }_0}\frac{x}{(a^2+x^2{)}^{{\displaystyle \frac{3}{2}}}} \\ so{\int }_{x=\infty }^{x=0}\frac{Q}{4\pi {\epsilon }_0}\frac{x}{(a^2+x^2{)}^{{\displaystyle \frac{3}{2}}}}dx(hereischangedthenotationfromltox) \\ herea\to radiusofthering \\ x\to positionfromthecenter \\ Q\to totalchargeonthesphere \\ letx=a\tan \left(\theta \right) \\ dx=ase{c}^2\left(\theta \right)d\theta sowhenx=0,\theta =0andx=\infty ,\theta ={90}^0 \\ {\int }_{x=\infty }^{x=0}E.dx={\int }_{x=\infty }^{x=0}-\frac{Q}{4\pi {\epsilon }_0}\frac{x}{(a^2+x^2{)}^{{\displaystyle \frac{3}{2}}}}dx={\int }_{x=\infty }^{x=0}-\frac{Q}{4\pi {\epsilon }_0}\frac{a\tan \left(\theta \right)}{(a^2+a^2{\tan }^2\left(\theta \right){)}^{{\displaystyle \frac{3}{2}}}}ase{c}^2\left(\theta \right)d\theta \\ =\frac{Q}{4\pi {\epsilon }_{0}a}{\int }_{\theta =90}^{\theta =0}-\sin \left(\theta \right)d\theta =\frac{Q}{4\pi {\epsilon }_{0}a}{\left[\cos \left(\theta \right)\right]}_{90}^0=\frac{Q}{4\pi {\epsilon }_{0}a} \\ putthevalueoftheparametersweget \\ {\int }_{x=\infty }^{x=0}E.dx=\frac{1.11\times {10}^{-10}\times 9\times {10}^9}{.5}=+2 \end{gathered}$$
Regard