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CBSE Class 12-science Answered

A parallel plate capacitor is charged by battery , which is 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) The energy stored in the capacitor
Asked by nitishkrnehu09 | 29 Dec, 2017, 08:10: PM
answered-by-expert Expert Answer
after disconnecting battery let us assume the charge on the plates is Q and the potential difference is V. The relation connecting capacitance C, charge Q and potential difference V is given by C = Q/V.
 
when dielectric material is inserted between parallel plates the capacitance increases and the new capacitance C' is given by C' = k×C, where k is dielectric constant. Charge Q remains same. Hence potential difference reduces and the new value of potential difference is V/k .
 
Energy stored in the capacitor before inserting dielectric  = (1/2)×C×V2
 
after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)2 = k×(1/2)×C×V2

Answered by | 31 Dec, 2017, 05:58: PM

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