A) \[30C\]
B) \[40C\]
C) \[10C\]
D) \[20C\]
Correct Answer: D
Solution :
: Let Q C be original charge of the capacitor and C be capacitance of the capacitor. \[\therefore \] Initial energy stored in the capacitor is \[U=\frac{1}{2}\frac{{{Q}^{2}}}{C}\] ?..(i) When an additional charge of 2 C is given to a capacitor, its charge on the capacitor becomes \[Q=Q+2\] ?.(ii) Now energy stored in the capacitor is \[U=\frac{1}{2}\frac{Q{{}^{2}}}{C}\] ??(iii) As per question \[U=U+\frac{21}{100}U=\frac{121}{100}U\] \[\therefore \] \[\frac{1}{2}\frac{Q{{}^{2}}}{C}=\frac{121}{100}\left( \frac{1}{2}\frac{{{Q}^{2}}}{C} \right)\](using (i) and (iii)) \[Q{{}^{2}}=\frac{121}{100}{{Q}^{2}}\] or \[Q=\frac{11}{10}Q\] \[Q+2=\frac{11}{10}Q\] (Using (ii)) \[10(Q+2)=11Q\] \[10Q+20=11Q\] or \[Q=20C\]
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