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CBSE Class 12-science Answered

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Asked by jain.pradeep | 29 Feb, 2020, 10:45: AM
answered-by-expert Expert Answer
(1) Electric field energystored in the capacitor = (1/2)CV2
 
Where C is equivalent capacitance C and V is potential difference across capacitor plates.
 
By introducing dielectric material of dielectric constant K, Capacitance increases by a factor of K .
 
But Capacitors are connected same battery, potential difference remains same.
 
Hence Electric field is increaed by a factor of K
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(2) Charge Q on capacitor C is given as,   Q = C V
 
when capacitance increases by K by introducing dielectric material of dielectric constant K, but potential difference remains same,
 
Charge on capacitor increases by a factor K
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(3) potential difference does not change , since same battery of constant EMF is connected acros  capacitors.
Answered by Thiyagarajan K | 29 Feb, 2020, 04:50: PM

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