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A900pF capacitor is charged by 100 V ...

A900pF capacitor is charged by 100 V sources .Calculate the electrostatic energy stored in the capacitor .The capacitor is then disconnected from the source and connected to another uncharged 900pF capacitor . Find the common potential of the system ?

Text Solution

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`C_I =900 pF =900 xx 10^(-12) F,V_1 =100 V,C_2 =900 V,C_2 =900pF =900 xx 10^(-12) F,V_2=0`
`q=CV =900 xx 10^(-12) xx 100= 9 xx 10^(-8) C `
The energy stored in the capacitor is
`U=1/2 CV^2 =1/2 qV =1/2 9 xx 10^(-8) xx 100`
` U=4.5 xx 10^(-6) J`
when the first capacitor is disconnected from the supply and connected to anther uncharged capacitor of capacitance `900^@ pF ` potential difference is
`V=(C_1 V_1 +C_2 V_2)/(C_1 +C_2)`
`=((900 xx 10^(-12)xx 100)+0)/(900 xx 10^(-12) +900 xx 10^(-12))`
`=(9 xx 10^(-8))/(1800 xx 10^(-12))`
` =(9xx 10^(-8))/(18 xx 10^(-10))`
`=0.5 xx 10^2`
`=50 V`
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