| A solid sphere having uniform charge density p and radius \[R\] is show in figure. A Spherical cavity of radius \[\frac{R}{2}\] is hollowed out. So the potential at \[o\]Will be (Assuming potential of infinity to be Zero) |
 |
A) \[\frac{11{{R}^{2}}\rho }{24{{\varepsilon }_{0}}}\]
B) \[\frac{5}{12}\frac{{{R}^{2}}\rho }{{{\varepsilon }_{0}}}\]
C) \[\frac{7\rho {{R}^{2}}}{12{{\varepsilon }_{0}}}\]
D) \[\frac{3}{2}\frac{{{R}^{2}}\rho }{{{\varepsilon }_{0}}}\]
Correct Answer: B
Solution :
| change of the whole sphere, \[Q=\left( \frac{4}{3}\pi {{R}^{3}} \right)\rho \] |
| Change of the city, \[q=\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{2}}\rho =\frac{Q}{8}\] |
| Now, \[{{V}_{o}}=\left[ {{\left( {{V}_{\operatorname{whole}\,sphere}} \right)}_{o}}-{{\left( {{V}_{eavity}} \right)}_{o}} \right]\] |
| \[=\frac{3}{8\pi {{\varepsilon }_{0}}}\frac{Q}{R}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\frac{Q}{8}}{\frac{R}{2}}\]\[=\frac{5}{16\pi {{\varepsilon }_{0}}R}\times \left( \frac{4}{3}\pi {{R}^{3}}\rho \right)\]\[=\frac{5}{12}\frac{{{R}^{2}}\rho }{{{\varepsilon }_{0}}}\] |
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