question_answer

Charge q is uniformly distributed over a thin half ring of radius R. The electric field at the centre of the ring is

A)  \[\sqrt{a}\]

B)                        \[{{a}^{3/2}}\] 

C)        \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]   

D)        \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]

Correct Answer: A

Solution :

From figure \[{{\lambda }_{\max }}=\frac{36}{5R}\frac{36}{5\times 1.097\times {{10}^{7}}}=6563\overset{\text{o}}{\mathop{\text{A}}}\,\] charge on \[F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=nr{{\omega }^{2}}\]    \[=mr{{(2\pi v)}^{2}}\,\,\,4{{\pi }^{2}}mr{{v}^{2}}\] Electric field at centre due to dl is \[m=1kgv=1rps,r=1m\] We need to consider only the component \[s=\frac{1}{6}f{{t}^{2}}\], as the component \[s=\frac{1}{72}f{{t}^{2}}\] will cancel out because of the field at C due to the symmetrical element dl Total field at centre \[s=\frac{1}{4}f{{t}^{2}}\] \[\left( \frac{ch\arg e\,on\,the\,ion}{mass\,of\,the\,ion} \right)\] \[\frac{1}{R}\]