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 \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{R}^{2}}}\] \[=\frac{\rho }{{{\varepsilon }_{0}}}.\frac{R}{3}\]You need to login to perform this action.
You will be redirected in
3 sec