Download the app

Questions  

A parallel plate capacitor having plate area ‘A’ and plate separation ‘d’ is joined to a battery of emf ‘V’ and internal resistance ‘R’ at  t=0 . Consider a plane surface of area A/3,  parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. [ The charge on the capacitor at time ‘t’ is given by q=CV1etτwhere  τ=CR ]

a
id=V2Re−td∈0AR
b
id=V3Re−td∈0AR
c
id=2V3Re−2td∈0AR
d
id=3V2Re−3td∈0AR

detailed solution

Correct option is B

Surface charge density,  σ=qA=CVA1−e−tτ Electric field b/w the plates of capacitor, E=σ∈0=CV∈0A1−e−tτ Electric flux from the given area, ϕE=EA3=CV3∈01−e−tτ Displacement current, id=∈0dϕEdt=∈0ddtCV3∈01−e−tτ=CV3τe−tτ ⇒id=V3Re−tCR ⇒id=V3Re−td∈0AR

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. if the plate area is 60 cm2, then the value of displacement current for 106sec  will be


phone icon
whats app icon