Two metal plates are separated by a distance d in a parallel plate condenser. A metal plate of thickness t and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :
d−td
(t−td)
(2−td)
1(1−td)
A
(t−td)
B
d−td
C
1(1−td)
D
(2−td)
Open in App
Solution
Verified by Toppr
The capacitance after inserting dielectric slab of thickness t and dielectric constant K is: ϵ0Ad−t(1−1K)
Was this answer helpful?
6
Similar Questions
Q1
Two metal plates are separated by a distance d in a parallel plate condenser. A metal plate of thickness t and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :
View Solution
Q2
Two metal plates form a parallel plate condenser. The distance between the plates is d. A metal plate of thickness d2 and of the same area is inserted completely between the plates. The ratio of capacitance's in the two cases (later to initial)
View Solution
Q3
A metallic plate of thickness (t) and face area of one side (A) is inserted between the plates of a parallel plate air capacitor with a separation (d) and face area (A). Then the equivalent capacitance is :
View Solution
Q4
Capacitance of a parallel plate capacitor becomes 43 times its original value if a dielectric slab of thickness t=d2 is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is
View Solution
Q5
Separation between the plates of a parallel plate capacitor is and the area of each plate is A. When a slab of material of dielectric constant and thickness t(t<d) is introduced between the plates, its capacitance becomes