A) \[0.6\,\,J\]
B) \[0.06\,\,J\]
C) \[0.03\,\,J\]
D) \[0.3\,\,J\]
Correct Answer: C
Solution :
The total work done in charging the capacitor \[\left( C \right)\] from the uncharged state to the final charge \[q\] will be equal to energy stored in the capacitor, given by \[U=\frac{1}{2}\frac{{{q}^{2}}}{C}=\frac{1}{2}C{{V}^{2}}\] Where \[q=CV\] and \[V\] is potential difference. Given,\[C=6\times {{10}^{-6}}\,\,F,\,V=100\] volt \[\therefore \] \[U=\frac{1}{2}\times 6\times {{10}^{-6}}\times {{\left( 100 \right)}^{2}}\] \[=3\times {{10}^{-2}}\,\,J\] \[U=0.03\,\,J\] Note: This energy resides in the electric field created between the plates of the charged capacitor.
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