question_answer

A solid non-conducting hemisphere of radius R has a uniformly distributed positive charge of volume density\[\rho \]. A negatively charged particle having charge q is transferred from centre of its base to infinity. Work performed in the process is

A) \[\frac{q\rho {{R}^{2}}}{4{{\varepsilon }_{0}}}\]    

B) \[\frac{q\rho {{R}^{2}}}{2{{\varepsilon }_{0}}}\]

C) \[\frac{q\rho {{R}^{2}}}{8{{\varepsilon }_{0}}}\]    

D) \[\frac{-q\rho {{R}^{2}}}{8{{\varepsilon }_{0}}}\]

Correct Answer: A

Solution :

[a] To calculate initial potential energy of the particle, first a thin hemispherical shell of radius x and radial thickness dx is considered. Volume of material of the shell \[=2\pi {{x}^{2}}\,dx\] charge on shell is \[dQ=\rho (2\pi {{x}^{2}}dx)\] Potential energy of the particle, due to charge of the shell considered is \[dU=\frac{-qdQ}{4\pi {{\varepsilon }_{0}}x}=\frac{-q\rho xdx}{4{{\varepsilon }_{0}}}\] \[\Rightarrow \,\,U=\int\limits_{x=0}^{x=R}{\frac{-q\rho xdx}{2{{\varepsilon }_{0}}}}=\frac{-q\rho {{R}^{2}}}{4{{\varepsilon }_{0}}}\] Finally potential energy is zero. Thus, work done \[\Delta U=\frac{q\rho {{R}^{2}}}{4{{\varepsilon }_{0}}}\]