1.

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q_(0),V_(0),E_(0) and U_(0) respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q,V,E

Answer»

`QgtQ_(0)`
`VgtV_(0)`
`EgtE_(0)`
`UgtU_(0)`

Solution :When electric stab is introduced capacity GETS increased while potential DIFFERENCE remains unchanged
`V=V_0,CgtC_0 ,Q=CV therefore QgtQ_0`
`U=1/2CV^(2) therefore ugtU_0,E=v/d` but v and d both are unchanged .
Therefore ,`E=E_0`, Therefore correct OPTION are (A) and (D).


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