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                                                                                          [Roorkee 1999]

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

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

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

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

Correct Answer: B

Solution :

                   From figure \[dl=R\text{ }d\theta \]; Charge on \[dl=\lambda R\text{ }d\theta \]     \[\left\{ \lambda =\frac{q}{\pi R} \right\}\] Electric field at centre due to dl is \[dE=k.\frac{\lambda Rd\theta }{{{R}^{2}}}\]. We need to consider only the component \[dE\,\cos \theta ,\] as the component dE sinq will cancel out because of the field at C due to the symmetrical element dl¢. Total field at centre \[=2\int_{\,0}^{\,\pi /2}{\,dE\cos \theta }\] \[=\frac{2k\lambda }{R}\int_{\,0}^{\,\pi /2}{\,\cos \theta \,d\theta }=\frac{2k\lambda }{R}=\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\] Alternate method : As we know that electric field due to a finite length charged wire on it's perpendicular bisector is given by \[E=\frac{2k\lambda }{R}\sin \theta .\] If it is bent in the form of a semicircle then \[\theta =\text{ 9}0{}^\circ \] Þ \[E=\frac{2k\lambda }{R}\] = \[2\times \frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{q/\pi R}{R} \right)\] = \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]