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The capacitor is disconnected from the battery and connected to another $900\;pF$ capacitor [Fig]. What is the electrostatic energy stored by the system?

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In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential.
Let the common potential difference be V′.
The charge on each capacitor is then $Q′ = CV'$. By charge conservation, $Q'=Q/2$.
This implies $V′ = V/2$.
The total energy of the system is $= U_1 + U_2$
$\qquad= 2 \times \large \frac{1}{2}$$Q' V$
$\qquad= \large\frac{1}{2} . \frac{1}{2} $$QV$
$\qquad = \large\frac{1}{2} $$ \times 4.5 \times 10^{-6}$
$\qquad=2.25 \times 10^{-6} J$
answered Jun 19, 2014 by meena.p

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