question_answer

If an insulated non-conducting sphere of radius \[R\]has charge density\[\rho \], the electric field at a point just on the surface of Sphere is:                                                                                               [BHU M-2003]

A)  \[\frac{\rho R}{3{{\varepsilon }_{0}}}\]                                

B)  \[\frac{\rho r}{{{\varepsilon }_{0}}}\]

C)  \[\frac{\rho r}{3{{\varepsilon }_{0}}}\]                                 

D)  \[\frac{3\rho R}{{{\varepsilon }_{0}}}\]

Correct Answer: A

Solution :

                 Key Idea: Since the sphere is non-conducting charge will be distribution throughout the volume of the sphere. When \[r\] is the volume charge density, then                                                 \[q=\frac{4}{3}\pi \,{{R}^{2}}\rho \] The electric field intensity due to a uniformly charged sphere at an external point is the same as if the entire charge on it were concentrated at the center of the sphere                                 \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\]                             \[\left( r>R \right)\]                 \[\therefore \]  \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{4}{3}\frac{\pi \,{{R}^{3}}\,\rho }{{{r}^{2}}}\]                                 \[=\frac{\rho }{{{\varepsilon }_{0}}}.\frac{{{R}^{3}}}{3{{r}^{2}}}\]                              \[\left( for\,\,r>R \right)\] If the point \[P\] less just on the surface of the sphere then r=R. Then                      \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{R}^{2}}}\]                                                      \[=\frac{\rho }{{{\varepsilon }_{0}}}.\frac{R}{3}\]