Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
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?
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 the electric field between the plates. Explain the origin of the factor `1/2`.
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 ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].
A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
In a parallel plate capacitor, the capacity increases if ______.
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
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.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
