CBSE Class 12-science Answered
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Asked by jain.pradeep | 18 Feb, 2020, 08:20: PM
Expert Answer
As shown in figure a battery of EMF E is connected a parallel plate capacitor of capacitance C through a resistance R.
Resistance R may be a connected resistance to control the current or resistance due to connecting wire.
if i is the charging current at an instant t, and q is the charge on the capacitor plates at same instant t
then we have , E = i R + ( q/C )
if we substitute i = dq/dt in above equation, we get , E = R (dq/dt) + (q/C) ..................(1)
Above equation can be re arranged as 
......................(2)
......................(2)Using the initial condition at t = 0, q = 0 and by integration we get the final charge Q on capacitor as
Q = EC [ 1 - e-t/(RC) ] ..................(3)
where RC is called as time constant of the charging circuit
--------------------------------------
when a capacitor of capacitance C is charged by a potential difference V, then charge Q = CV
Capacitance C of air filled parallel plate capacitor is given by, C = [ εo A ] /d
where A is area of plates , d is the separation distance between plates and εo is the permittivity of free space .
After disconnecting the potential difference, If dielectric material of dielectric constant k is introduced between parallel plates
and the separation distance d is doubled then
changed capacitance C' = k [ εo A ] / (2d) = (k/2) C
After disconnecting the potential difference which is the source of charging the capacitor, Charge Q will remain same and capacitance is changed due to chane in distance and presence of dieclectric material.
Hence potential difference V' between parallel plates is given by
V' = Q/C' = (2/k) Q/C = (2/k) V
Answered by Thiyagarajan K | 19 Feb, 2020, 10:32: AM
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