Q.

In the circuit diagram a capacitor which is initially uncharged is connected to an ideal cell of emf ε through a resistor R and switch S. A leaky dielectric fills the space between the plates of capacitor. The capacitance of the capacitor with dielectric is C. Resistance of dielectric is R'=R. If switch is closed at t=0, then Charge on capacitor at t = RC ln 2 is
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a

Cε4

b

Cε2

c

Cε8

d

3Cε8

answer is D.

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Detailed Solution

 Co=oAd  C=KoAd  εiRqc=0,   dqdt=iic=i(qCR) 
 ε[qc+Rdqdt]qc=0q=12cε(1e2t/Rc) 
=12cε(1e2ln2)=12cε(1eln4)=38cε 

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