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 × 10−3m2 and the separation between the plates is 3 mm.
- Calculate the capacitance of the capacitor.
- If this capacitor is connected to 100 V supply, what would be the charge on each plate?
- How would charge on the plates be affected, if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
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/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 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 is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
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?
Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
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?
