question_answer

A thin glass rod is bent into a semi circle of radius r. A charge + q is uniformly distributed along the upper half and a charge - q is uniformly distributed along the lower half, as shown in the figure. The magnitude and direction of the electric field E produced at P, the centre of the circle, will be

A)  0

B)  \[\frac{q}{{{\varepsilon }_{0}}{{\pi }^{2}}{{r}^{2}}}\] perpendicular to the line OP and directed downward

C)  \[\frac{q}{{{\varepsilon }_{0}}\pi {{r}^{2}}}\] perpendicular to the line OP and directed downward

D)  \[\frac{q}{{{\varepsilon }_{0}}\pi {{r}^{2}}}\] along the axis OP

Correct Answer: B

Solution :

                \[E=2\int{dE\cos \theta }\]                 \[E=\frac{2}{4\pi {{\varepsilon }_{0}}}\frac{2q}{\pi {{R}^{3}}}\int_{o}^{\pi /2}{\cos \theta \,d\theta }\]                 \[=\frac{q}{{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]                 \[=\frac{q}{{{\varepsilon }_{0}}{{\pi }^{2}}{{R}^{2}}}\] Perpendicular to the line OP and directed downward.