0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t (t < d) is introduced between the plates, its capacitance becomes
  1. ϵ0Ad+t⎪ ⎪ ⎪11k⎪ ⎪ ⎪
  2. ϵ0Ad+t⎪ ⎪ ⎪1+1k⎪ ⎪ ⎪
  3. ϵ0Adt⎪ ⎪ ⎪11k⎪ ⎪ ⎪
  4. ϵ0Adt⎪ ⎪ ⎪1+1k⎪ ⎪ ⎪

A
ϵ0Ad+t⎪ ⎪ ⎪11k⎪ ⎪ ⎪
B
ϵ0Adt⎪ ⎪ ⎪11k⎪ ⎪ ⎪
C
ϵ0Adt⎪ ⎪ ⎪1+1k⎪ ⎪ ⎪
D
ϵ0Ad+t⎪ ⎪ ⎪1+1k⎪ ⎪ ⎪
Solution
Verified by Toppr

Was this answer helpful?
11
Similar Questions
Q1
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
View Solution
Q2
Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t (t < d) is introduced between the plates, its capacitance becomes
View Solution
Q3
Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t<d) is introduced between the plates, its capacitance becomes




View Solution
Q4
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
View Solution
Q5
Distance between the plates of a parallel plate capacitor is d and area of each plate is A. When a slab of dielectric constant K and thickness t is placed between the plates, its capacity becomes
View Solution