AIPMT | 2015CapacitanceHard
Question
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?
Options
A.The energy stored in the capacitor decreases K times.
B.The change in energy stored is 
C.The charge on the capacitor is not conserved.
D.The potential difference between the plates decreases K times.
Solution
Once the capacitor is charged, its charge will be
constant Q = CV
When dielectric slab is inserted
CNew = KC
E =
⇒ ENew =
Einitial
V =
so VNew =
V
Hence option (3)
constant Q = CV
When dielectric slab is inserted
CNew = KC
E =
V =
Hence option (3)
Create a free account to view solution
View Solution FreeMore Capacitance Questions
In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then inthe steady ...The two parallel plates of a condenser have been connected to a battery of 300 V and the charge collected at each plate ...The distance between the plates of a parallel plate condenser is d. If a copper plate of same area but thickness d/2 is ...The area of the plates of a parallel plate condenser is A and the distance between the plates is 10 mm. There are two di...Two spheres of radii 1 cm and 2 cm have been charged with 1.5 × 10−8 and 0.3 × 10−7 coulomb of pos...