Electrostatics Question 738
Question: 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?
Options:
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 $
Show Answer
Answer:
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}{{{( 2R )}^{2}}}-E $
$ =\frac{1}{4\pi {\varepsilon _{0}}}\frac{\rho ( 4/3 )\pi R^{3}}{4R^{2}}-E $
$ =\frac{\rho R}{12{\varepsilon _{0}}}-E $
