the plates of parallel plate air capacitor are connected to a battery of emf 6 volt. if the space between the plates is filled with a medium of dielectric constant 2 with battery still in connection . the new potential difference would be ???
options:
a)6v
b)3v
c)12v
d)18v
(Ans must be with solution)
Answers
Given :
EMF of battery = 6Volts
Dielectric constant of medium = 2
To Find :
The new potential difference across the parallel plates after filling the space between the plates with a dielectric material.
Solution :
Let capacitance of capacitor be C.
- Charge = Qo
- Voltage = Vo
After filling the space between parallel plates with dielectric material (with battery connected)
- New charge on the plates will be KQ｡
- New capacitance of the capacitor will be KC｡
As we know that capacitance of the capacitor is measured as the ratio of charge on plate to the pd across plates.
Mathematically, C = Q/V
By substituting the values, we get
➠ C = Q/V
➠ KC｡= KQ｡/V
➠ V = Q｡/C｡
In the first case, C｡ = Q｡/V｡
➠ V = V｡= 6 Volts
∴ Potential difference across parallel plates will remain constant
■ Remember :
Effect of dielectric with battery connected across the capacitor,
- Q = KQ｡
- V = V｡
- E = E｡
- C = KC｡
- U = KU｡
Effect of dielectric with battery disconnected from the capacitor,
- Q = Q｡
- V = V｡/K
- E = E｡/K
- C = KC｡
- U = U｡/K