A) \[\frac{\rho {{R}^{2}}{{q}_{0}}}{3{{\varepsilon }_{0}}}\]
B) \[\frac{\rho {{R}^{2}}{{q}_{0}}}{6{{\varepsilon }_{0}}}\]
C) \[\frac{\rho {{R}^{2}}{{q}_{0}}}{4{{\varepsilon }_{0}}}\]
D) \[\frac{\rho {{R}^{2}}{{q}_{0}}}{12{{\varepsilon }_{0}}}\]
Correct Answer: D
Solution :
The electric field inside the cavity is constant and is equal to the field at point C and it is same as there were no cavity. \[\Rightarrow E=\frac{\rho R/2}{3\,{{\varepsilon }_{0}}}=\frac{\rho R}{6\,{{\varepsilon }_{0}}}\]radially outward .Kinetic energy , change = work done \[={{q}_{0}}\Delta V={{q}_{0}}ER/2\] \[\Rightarrow {{V}_{C}}-{{V}_{A}}=E\frac{R}{2}=\frac{\rho {{R}^{2}}}{12{{\varepsilon }_{0}}}\] \[\therefore K.E.={{q}_{0}}({{V}_{C}}-{{V}_{A}})=\frac{\rho {{R}^{2}}{{q}_{0}}}{12{{\varepsilon }_{0}}}\]
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