Q. 64.0( 1 Vote )

A parallel-plate A. The battery will supply more charge.

B. The capacitance will increase.

C. The potential difference between the plates will increase.

D. Equal and opposite charges will appear on the two faces of the metal plate.

Answer :

Let E0=V0/d be the electric field between the plates when there is no electric and the potential difference is V0. If the dielectric of dielectric constant K is now inserted , the electric field in the dielectric will be

The potential difference will then be


t=thickness of dielectric slab

d=separation between the plates of capacitor

Here, since metal plate is of negligible thickness , t=0

And V=E0d=V0

Thus the potential remains same c) is incorrect) and the charge Q0 on plates also remains same.


capacitance remains same.b) is incorrect)

Since charge on the capacitor remains same, no extra charge is supplied by the batterya) is incorrect)

We know that when dielectric is introduced between the plates of capacitor this polarized dielectric is equivalent to two charged surfaces with induced surface charges Q’ and -Q’

Here, the dielectric is the metal plate and therefore equal and opposite charges appear on the two faces of metal plate.

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