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A parallel plate capacitor is charged by...

A parallel plate capacitor is charged by a battery after some time, the battery is disconnected and a dielectric slab with its thicnkess equal to the plate so reparation is insected between the plates. How will (i) the capacitance of the capacitor (ii) potential difference between the plates & (iii) the energy sotred in the capacitor the affected ?
`Q_(0)` -charge `V_(0)`- potential difference, `C_(0)`- capacitance, `E_(0)` electric field.
`U_(0)`- energy spred, before the dielectric slab is inserted.
`Q_(0)=C_(0) V_(0), (V_(0))/(d), U_(0)=(1)/(2)C_(0)V_(0)^(2)?`

Text Solution

Verified by Experts

(i) When battery is disconnected, the capacitance increases from `C_(0)` to C
`C= epsi_(r) C_(0)`
(ii) New potential difference between the plates
`V=(Q_(0))/(C)=(C_(0)V_(0))/(C)=(C_(0)V_(0))/(epsi_(0)C_(0))=(V_(0))/(epsi_(r))`
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