The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4cm is 2μF. If the separation is reduced to half and it is filled with a dielectric substance of value 2.8, then the final capacity of the capacitor is
15.6μF
22.4μF
19.2μF
11.2μF
A
19.2μF
B
11.2μF
C
15.6μF
D
22.4μF
Open in App
Solution
Verified by Toppr
Given, C=ε0Ad=2μF and C′=Kε0Ad′=2.8ε0Ad/2=5.6ε0Ad ⇒C′=5.60×2=11.2μF
Was this answer helpful?
0
Similar Questions
Q1
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4cm is 2μF. If the separation is reduced to half and it is filled with a dielectric substance of value 2.8, then the final capacity of the capacitor is
View Solution
Q2
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2μF. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is
View Solution
Q3
A parallel plate capacitor with air has a capacity of 8pF. Calculate the capacity of the capacitor if the distance between the plates in halved and the space between them is fully filled with a substance of dielectric constant 5.
View Solution
Q4
An air-filled parallel plate capacitor has a capacity 2pF. The separation between the plates is doubled and the interspace is filled with wax, If the capacity is increased to 6 pF, the dielectric constant of the wax is :
View Solution
Q5
The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant 3, the capacity of the condenser becomes: