| Directions: (26-30) |
| Spherical Capacitor |
| The electrical capacitance of a conductor is the measure of its ability to hold electric charge. |
| An isolated spherical conductor of radius R. The charge Q is uniformly distributed over its entire surface. It can be assumed to be concentrated at the centre of the sphere. |
| The potential at any point on the surface of the spherical conductor will be\[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}\]. |
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| Capacitance of the spherical conductor situated in vacuum \[C=\frac{q}{V}=\frac{Q}{\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}}\] or \[C=4\pi {{\varepsilon }_{0}}R\] |
| Clearly, the capacitance of a spherical conductor is proportional to its radius. |
| The radius of the spherical conductor of 1 F capacitance is \[R=\frac{1}{4\pi {{\varepsilon }_{0}}}\]. C and this radius is about 1500 times the radius of the earth\[\left( \tilde{\ }6\times {{10}^{3}}\,km \right)\]. |
A) 90 cm
B) 45 cm
C) 45 m
D) 90m
Correct Answer: B
Solution :
Here \[C=50\,pF=50\times {{10}^{-12}}F,\,V={{10}^{4}}V\] \[R=\frac{1}{4\pi {{\varepsilon }_{0}}}.\,C=9\times {{10}^{9}}\,m{{F}^{-1}}\times 50\times {{10}^{-12}}F\] \[=45\times {{10}^{-2}}m=45\,cm\]
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