Advertisements
Advertisements
Question
A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.
Advertisements
Solution
Electric field strength for a parallel plate capacitor,
`E = Q/(∈_0A)`
Electric flux linked with the area,
`phi_E = EA = Q/(∈_0A) xx A/2 = Q/(2∈_0)`
Displacement current ,
`I_d = ∈_0 (dphi_E)/dt = ∈_0 d/dt(Q/(∈_0 2))`
`I_d = 1/2((dQ)/dt)` ....(i)
Charge on the capacitor as a function of time during charging,
`Q = εC[1 - e^(-t"/"RC)]`:
Putting this in equation (i), we get :
`I_d = 1/2 εC d/dt(1 - e^(-t"/"RC))`
`I_d = 1/2 εC(-e^(-t"/"RC)) xx (-1/(RC))`
`C = (A∈_0)/d`
⇒ `I_d = ε/(2R) xx e^(-(td)/(ε_0AR))`
APPEARS IN
RELATED QUESTIONS
A parallel plate capacitor (Figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.
- What is the rms value of the conduction current?
- Is the conduction current equal to the displacement current?
- Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current?
Without the concept of displacement current it is not possible to correctly apply Ampere’s law on a path parallel to the plates of parallel plate capacitor having capacitance C in ______.
Displacement current is given by ______.
The displacement of a particle from its mean position is given by x = 4 sin (10πt + 1.5π) cos (10πt + 1.5π). The motion of the particle is
Displacement current goes through the gap between the plantes of a capacitors. When the charge of the capacitor:-
Which of the following is the unit of displacement current?
A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. What is the value of the displacement current for 10-6 second if the plate area is 60 cm2?
According to Maxwell's hypothesis, a changing electric field gives rise to ______.
A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V = V0 sinωt The displacement current between the plates of the capacitor would then be given by ______.
An electromagnetic wave travelling along z-axis is given as: E = E0 cos (kz – ωt.). Choose the correct options from the following;
- The associated magnetic field is given as `B = 1/c hatk xx E = 1/ω (hatk xx E)`.
- The electromagnetic field can be written in terms of the associated magnetic field as `E = c(B xx hatk)`.
- `hatk.E = 0, hatk.B` = 0.
- `hatk xx E = 0, hatk xx B` = 0.
Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is `(ε_0mu_r)/2 (dE)/(dt)` (symbols having usual meaning).
AC voltage V(t) = 20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is ______.
[take ε0 = 8.85 × 10-12 F/m]
A particle is moving with speed v = b`sqrtx` along positive x-axis. Calculate the speed of the particle at time t = τ (assume that the particle is at origin at t = 0).
A parallel plate capacitor is charged to 100 × 10-6 C. Due to radiations, falling from a radiating source, the plate loses charge at the rate of 2 × 10-7 Cs-1. The magnitude of displacement current is ______.
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its palates is increasing at a constant rate with time. The magnetic field arising due to displacement current is ______.
