question_answer

Figure shows a uniformly charged hemisphere of radius R. It has a volume charge density \[\rho .\] If the electric field at a point 2R, above its center is E, then what is the electric field at the point 2R below its center?

A) \[\rho R/6{{\varepsilon }_{0}}+E\]         

B)        \[\rho R/12{{\varepsilon }_{0}}-E\]

C) \[-\rho R/6{{\varepsilon }_{0}}+E\]

D)        \[\rho R/12{{\varepsilon }_{0}}+E\]

Correct Answer: B

Solution :

[b] Let us complete the sphere. Electric field due to upper part at B = E (given) Electric field B = electric field due to fall sphere electric field due to upper part at B=electric field due to full sphere -electric field due to upper part \[=\frac{kQ}{{{\left( 2R \right)}^{2}}}-E\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\rho \left( 4/3 \right)\pi {{R}^{3}}}{4{{R}^{2}}}-E\] \[=\frac{\rho R}{12{{\varepsilon }_{0}}}-E\]