A) 0.2J
B) 06 J
C) 1.2J
D) 2.4J
Correct Answer: B
Solution :
(b) 0,6 J Here, \[r=18cm=18\times {{10}^{-2}}m,\,q=5\times {{10}^{-6}}C\] As \[C=4\pi {{\varepsilon }_{0}}r\] \[=\frac{18\times {{10}^{-2}}}{9\times {{10}^{9}}}=2\times {{10}^{-11}}F\] Energy of charged conductor is \[U=\frac{{{q}^{2}}}{2C}=\frac{{{\left( 5\times {{10}^{-6}} \right)}^{2}}C}{2\times 2\times {{10}^{-11}}F}=0.625J\]You need to login to perform this action.
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