Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged?
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 ______.
If the total energy of a particle executing SHM is E, then the potential energy V and the kinetic energy K of the particle in terms of E when its displacement is half of its amplitude will be ______.
A spring balance has a scale that reads 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this spring, when displaced and released, oscillates with a period of 0.60 s. What is the weight of the body?
A cylinder of radius R, length Land density p floats upright in a fluid of density p0. The cylinder is given a gentle downward push as a result of which there is a vertical displacement of size x; it is then released; the time period of resulting (undampe (D) oscillations is ______.
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?
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.
A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?
You are given a 2 µF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
Sea water at frequency ν = 4 × 108 Hz has permittivity ε ≈ 80 εo, permeability µ ≈ µo and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source V(t) = Vo sin (2πνt). What fraction of the conduction current density is the displacement current density?
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 ______.
Draw a neat labelled diagram of displacement current in the space between the plates of the capacitor.
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, there is no conduction current:
