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a 10 uF capacitpr is connected to a 12 V battery till its fully charged. it is then disconnected and to an uncharged 20 uF capacitor.find energy stored on 20 uF capaciotr when it has reached an electro static equilibrium

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Answer:

Explanation:

The charge on 10μF capacitor =  10 x 12 x 10⁻⁶ = 120 μC

when it is connected with 20μF capacitor both acquires common potential whose value is

= 120 x 10⁻⁶ /( 10 +20) x 10⁻⁶ = 4 V.

Energy stored in 20μF capacitor =1/2 x 20 x 10⁻⁶ x 4 x 4 = 160 x 10⁻⁶ J.

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