A 0.1 F capacitor is charged by a 10 V battery. After disconnecting the battery, this charged capacitor is connected with an uncharged capacitor. If the charge is equally shared between the two, then what will be the energy stored in the two capacitors? Compare this energy with the energy stored initially in the first capacitor.

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Solution

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$I n i t i a l c h a r g e o n t h e c a p a c i t o r i s q_{1} = C \times V q_{1} = 0.1 \times 10 = 1 C e n e r g y s t o r e d i n t h e c a p a c i t o r i s E_{i} = \frac{{q_{1}}^{2}}{2 C} = \frac{1}{2 \times 0.1} = 5 J W h e n c o n n e c t e d w i t h o t h e r c a p a c i t o r c h a r g e i s e q u a l l y d i s t r i b u t e d i t m e a n s c a p a c i t o r o f t h e o t h e r c a p a c i t o r i s a l s o 0.1 F F i n a l c h a r g e o n b o t h t h e c a p a c i t o r i s q_{2} = \frac{1}{2} C f i n a l e n e r g y o f b o t h t h e c a p a c i t o r s a r e e q u a l E_{f} = \frac{{q^{2}}_{2}}{2 \times C} = \frac{1}{4 \times 2 \times 0.1} = \frac{5}{4} = 1.25 J T o t a l f i n a l e n e r g y = 2 \times 1.25 = 2.5 J r a t i o o f f i n a l e n e r g y o f \sin g l e c a p a c i t o r t o t h a t o f t h e i n i t i a l e n e r g y o f t h e c a p a c i t o r i s \frac{E_{f}}{E_{i}} = \frac{1.25}{5} = 0.25 R e g a r d s$

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