Narendran
Last Activity: 8 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.