# Two identical capacitors are connected in series.charge on each capacitor is q.a dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap,the battery of 2V remains connected.find the charge on each capacitor

Narendran
34 Points
7 years ago
Hello Friend.
*Let the dielectric slab with dielectric constant ‘K’ be filling the 1’st capacitor,
C1(new)=KC1.
As Q=CV ….. V=2v remains constant
there for Q1(new)=’Kq’.
*For the other capacitor charge remains ‘q’ itself,
as the permeability remains unchanged.

Narendran
34 Points
7 years ago
I’am really sorry there is probably some ambiguity in my previous answer this question is quiet tricky please do check solution for this from other resources.
The charge would probably increase on both(consider capacitors being alike and check).
Sorry again.
Narendran
34 Points
7 years ago
For C1
Capacitance is constant; Charge increases; Potential
difference increases and Stored energy increases.
For C2
Capacitance increases; Charge increases; Potential difference decreases and Stored
energy decreases.
[Explanation: The initial capacitance of the combination is
1/CI=1/C1+1/C2=1/C+1/C=2/C;
Ci=C/2.

Both have equal charge Q and the potential difference across each of the capacitors
is V/2.

If the dielectric is slipped between the plates of the capacitor C2, with the battery
still connected, its capacitance increases

The final capacitance of the circuit,

1/Cf=1/C1f+1/C2f=1/C+1/KC; Cf=C(K/(K+1).
C/2

The total capacitance of the circuit increases.
Since V is constant, charge increases in both capacitors.

Qf=C(K/(K+1)V   …..Put V=2 volt here.