Home
Class 12
PHYSICS
The charge on a parallel plate capacitor...

The charge on a parallel plate capacitor varies as `q=q_(0)cos 2pi v t`. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through th capacitor.

Text Solution

Verified by Experts

Displacement current in capacitor,
`I_(d)=I_(c )=(dq)/(dt) " "` ….(1)
Here, `q=q_(0)cos 2pi vt`
Substituting in equation (1),
`I_(d)=I_(c )=(d)/(dt)(q_(0)cos 2pi vt)`
`therefore I_(d)=I_(c )=-q_(0)sin 2pi vt xx 2pi v`
`therefore I_(d)=I_(c )=-2pi vq_(0)sin 2pi vt`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROMAGNETIC WAVES

    KUMAR PRAKASHAN|Exercise SECTION C NCERT EXEMPLAR SOLUTION (Short Answer Type Questions)|9 Videos

  • ELECTROMAGNETIC WAVES

    KUMAR PRAKASHAN|Exercise SECTION C NCERT EXEMPLAR SOLUTION (Long Answer Type Questions)|11 Videos

  • ELECTROMAGNETIC WAVES

    KUMAR PRAKASHAN|Exercise SECTION C NCERT EXEMPLAR SOLUTION (Multiple Choice Questions)|13 Videos

  • ELECTROMAGNETIC INDUCTION

    KUMAR PRAKASHAN|Exercise Section D MCQs (MCQs asked in Competitive Exams )|38 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    KUMAR PRAKASHAN|Exercise Section D ( MCQs asked in Cometitive Exams )|39 Videos

Similar Questions

Explore conceptually related problems

Sea water at frequency v=4xx10^(8) Hz has permittivity in ~~ 80 in_(0) , permeability mu = mu_(0) and resistivity rho = 0.25 Omega , Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t)=V_(0)sin 2pi vt . What fraction of the conduction current density is the displacement current current density ?

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1//2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1//2 .

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric Held between the plates. Explain the origin of the factor 1//2

An ideal cell is connected across a capacitor as shown in figure. The initial separation between the plates of a parallel plate capacitor is d. The lower plate is pulled down with a uniform velocity v. Neglect the resistance of the cirucit. Then the variation of charge on capacitor with time is given by

The plates of a parallel plate capacitor have an area of 90 cm^2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

The plates or a parallel plate capacitor have an area of 90 cm^(2) each and are separated by 2.5 mm. The capacitor is charged by connecting It to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor ? (b) View this energy as stored in the electrostatic field between the plates, und obtain the energy per unit volume "· Hence arrive at a relation between u and the magnitude of electric field E between the plates.

What is the area of the plates of a 2 F parallel plate capacitor given that the separation between the plates is 0.5 cm ?

The plates of small size of a parallel plate capacitor are charged as shown. The force on the charged particle of 'q' at a distance 'l' from the capacitor is : (Assume that the distance between the plates is d ltlt l )

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

An electron passes between two parallel plate of a capacitor as shown in the diagram. The electron enters the plate at t=0 and it comes out of the plates at time t=T. The electron hits the screen at A, as shown. Now a time varying potential difference V is applied between the plates, as shown in the graph. (Neglect the gravity). The point where the electron now will strike (as compared to point A) will be

Home
Profile