A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of:
(i) capacitance
(ii) potential difference between the plates
(iii) electric field between the plates
(iv) the energy stored in the capacitor .
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Answer:
Let the original capacitance, potential difference be C and V respectively.
then the new capacitance will be C'=Ck (where k is the dielectric value).
The new Potential Difference will be V'=V/k.
And the energy stored in the capacitor will be=((Ck))/2
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