Two parallel plate capacitors of capacitance C each are connected in series with a battery of emf `epsilon`. Then, one of the capacitors is filled with a dielectric of dielectric constane K.
i. Find the change in electric field in the two capacitors, if any
ii. What amount of charge flows through the battery?
iii Find the change in the energy stored in the circuit, if any.

Explanation

i. Two capacitors `A` and `B` initially have same charge `Q` and potential `V=Q//C`. The electric field between the capacitor plates is given by `E=V//d`. Since the two capacitors are connected series with the battery, the sum of patentials across the capacitors must be equal to `(epsilon)`,i.e.,
`epsilon=2V=(2Q)/C` (i)
ii. When one of the capacitors, say A, if filled will a dielectric, the capacity of A increases to `C'=KC`, that of B remains unchanged, i.,C. Suppose, charge on the capacitors becomes `(A)'` and potentials across `A` and `B` become `V_(A)` and `V'_(B)`, respectively, with `V'_(A)+V'_(B)=epsilon`. Hence, we have
`V'_(A)=(Q')/C'=(Q')/(KC)`
and `V'_(B)=(Q')/(C')`
Hence,
`epsilon=(Q')/C'(1+1/K)` (ii)
From Eqs. (i) and (ii) we get
`Q'=(2K)/(1+K)Q`
Since `Kgt1`,so `Q'gtQ`. Also
`V'_(A)=(2K)/(1+K)V`
and `V'_(B)=(2K)/(1+K)V`
Thus the electric field (or potential difference) in capacitor A increase by a factor of `2K//(1+K)` while that in B increases by ka factor of `2K//(1+K)`. The amount of charge that flows into the circuit is given by
`DeltaQ=Q'-Q((2K)/(1+K)-1)Q=(K-1)/(k+1)Q`
`1/2(K-1)/(K+1)Cepsilon`
iii. Initially, the energy is given by
`U_(i)=2xx1/2CV^(2)=CV^(2)`
Finally, the energy is
`U_(f)=1/2C'V+(A)^(2)+(1)/(2)CV'_(B)^(2)`
`=1/2KC(2/(1+K))^(2)V^(2)+1/2C((2K)/(1+K))^(2)V^(2)`
`=(4K)/(1+K)(1/2CV^(2))=(2K)/(1+K)U_(i)`