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A fully charged parallel plate capacitor...

A fully charged parallel plate capacitor is connected across an unchanged identical capacitor . Show that the energy stored in the combination is less than that stored initially in the single capacitor .

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Let a parallel plate capacitor of capacitance C is fully charged to a potential V so that electrostatic energy stored in it is
`U = (1)/(2) CV^(2)`
When this capacitor is connected across an unchanged identical capacitor , it shares its charge with that and common potential of two capacitors become :
`V. = (C_(1) V_(1) + C_(2) V_(2))/(C_(1) + C_(2)) = (CV + 0)/(C + C) = (V)/(2)`
`therefore` Electrostatic energy stored in the capacitor combination now is
`U. = (1)/(2) C. V^(2) = 1/2 (2 C) ((V)/(2))^(2) = (1)/(4) CV^(2)`
This shows that U. `lt U` . In fact `U. = (U)/(2)`
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