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In a circuit shown in fig find the pote...

In a circuit shown in fig find the potentail difference between the left and right plates of each capacitor.

Text Solution

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The correct Answer is:
`V_(1)=(q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2))V_(2)=(-q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2))`.
`varphi_(1)=(q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2)), varphi_(2)=(-q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2))`
Let `xi_(2) gt xi_(1)`, then using `-Delta varphi =0` in the closed circuit, (Fig)

`(-q)/(C_(1))+xi_(2)-(q)/(C_(2))-xi_(1)=0 q=((xi_(2)-xi_(1))C_(1)C_(2))/(C_(1)+C_(2))`
Hence the p.d across the left and right plate of capacitors,
`varphi_(1)=(q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2))` and similarly
`varphi_(2)=(-q)/(C_(1))=((xi_(2)-xi_(1))C_(2))/(C_(1)+C_(2))`.
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