question_answer

The energy stored in the capacitor as shown in the figure is \[4.5 \times 1{{0}^{-}}^{6}J\]. If the battery is replaced by another capacitor of 900 pF as shown in the figure (b), then the total energy of system is  

A) \[4.5\times {{10}^{-6}}J\]                    

B) \[2.25 \times  1{{0}^{-6}}\,J\]

C) Zero                             

D) \[9\times 1{{0}^{-}}^{6}\,J\]

Correct Answer: B

Solution :

\[\operatorname{Energy} stored in the capacitor =\frac{1}{2}\frac{{{Q}^{2}}}{C}\] \[\operatorname{Q}=\,\,CV=900\times {{10}^{-12}}F\times 100\,V\] \[\therefore \,\, Q=\,\,9\times {{10}^{-8}}\,C\] Energy of the capacitor when fully charged \[=\,\,\frac{1}{2}\frac{{{Q}^{2}}}{C}=4.5\times {{10}^{-\,6}}J\] The total charge is conserved. In figure (b), total capacitance \[= C = 2 \times  C = 2 \times  900 pF\] \[\therefore \,\,\,\,Final\,\,Energy=\,\,\frac{1}{2}\frac{{{Q}^{2}}}{C}=\frac{1}{2}.\frac{{{Q}^{2}}}{2C}\] \[\therefore \,\,\, Final Energy =\,\,\frac{4.5\times {{10}^{-\,6}}J}{2}= 2.25 \times  1{{0}^{-}}^{6}J\]