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In a parallel plate capacitor of plate a...

In a parallel plate capacitor of plate area A, plate separation d and charge Q, the force of attraction between the plates is F.

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Let the new charge on the parallel plate capacitor be `+q_(1)` and that on the concentric shells be `(q-q_(1))`, such that the potential differences across the shells would be same as the potential difference across the plates.
`phi_("innershell")=((q-q_(1)))/(4piepsilon_(0)(2d))+(-(q-q_(1)))/(4piepsilon_(0)d)=(-(q-q_(1)))/(4piepsilon_(0)(2d))`
`phi_("outershell")=(q-q_(1))/(4piepsilon_(0)(2d))+(-(q-q_(1)))/(4piepsilon_(0)(2d))=0`
Potential difference, `V_("shells")=phi_("outer")-phi_("inner")=((q-q_(1)))/(4piepsilon_(0)(2d))`
For parallel plate, `V_("plates")=(q_(1))/(4piepsilon_(0)d)`
`V_("shells")=V_("plates")`
`((q-q_(1)))/(4piepsilon_(0)(2d))=(q_(1))/(4piepsilon_(0)d)`
`implies (q-q_(1))=2q_(1)`
`q_(1)=q//3`
`therefore` Charge on the outer shell = `(q-q_(1))=(2)/(3)q`
Charge on the inner shell = `-2q//3`
Energy left in the parallel plate capacitor = `(1)/(2)(q_(1)^(2))/(C)=(q^(2))/(72piepsilon_(0)d)`
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