Advertisements
Advertisements
Question
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
Advertisements
Solution
Using Gauss’ law, the electric flux ΦE of a parallel plate capacitor having an area A, and a total charge Q is
`phi_E=EA=1/in_0Q/AxxA`
`=Q/in_0`
Where electric field is
`E=Q/(Ain_0)`
As the charge Q on the capacitor plates changes with time, so current is given by
i = dQ/dt
`:.(dphi_E)/dt=d/dt(Q/in_0)=1/in_0(dQ)/dt`
`=>in_0(dphi_E)/dt=(dQ)/dt=i`
This is the missing term in Ampere’s circuital law.
So the total current through the conductor is
i = Conduction current (ic) + Displacement current (id)
`:.i=i_c+i_d=i_c+in_0(dphi_E)/dt`
As Ampere’s circuital law is given by
`:.ointvecB.vec(dl)=mu_0I`
After modification we have Ampere−Maxwell law is given as
`ointB.dl=mu_0i_c+mu_0in_0(dphi_E)/dt`
The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and displacement current.
APPEARS IN
RELATED QUESTIONS
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
For a one dimensional electric field, the correct relation of E and potential V is _________.
In a parallel plate capacitor, the capacity increases if ______.
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:
- Key K is kept closed and plates of capacitors are moved apart using insulating handle.
- Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
- In A: Q remains same but C changes.
- In B: V remains same but C changes.
- In A: V remains same and hence Q changes.
- In B: Q remains same and hence V changes.
