Advertisements
Advertisements
प्रश्न
Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.
- Calculate the capacitance and the rate of charge of the potential difference between the plates.
- Obtain the displacement current across the plates.
- Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Advertisements
उत्तर
Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε0 = 8.854 × 10−12 C2 N−1 m−2
(a) Capacitance between the two plates is given by the relation,
C = `(ε_0"A")/"d"` .....[∵ A = πr2 = 3.14 × (0.12)2]
= `(8.854 xx 10^-12 xx 3.14 xx (0.12)^2)/0.05`
= 8.01 × 10−12 F
= 8.01 pF
Charge on each plate, q = CV ⇒ V = `"q"/"C"`
`therefore "dV"/"dt" = 1/"C" "dq"/"dt" = 1/"C" "I" ...(because "dq"/"dt" = "I")`
Rate of charge of the potential difference `"dV"/"dt"` = `0.15/(8.01 xx 10^-12)`
= 1.875 × 109 V s−1
(b) Displacement current across the plates,
`"I"_"d" = epsilon_0 ("d"phi_"E")/"dt"`
Where `phi_"E"` is the electric flux passing through a closed loop between the plates.
∵ The electric field E between the plates = `"q"/(epsilon_0 "A")`
∴ If the area of the loop is A then,
`phi_"E" = oint vec("E") * "d" vec("A") = oint "E dA" ....[because vec("E") ⊥ "d" vec("A")]`
`=> phi_"E" = "EA" = "q"/epsilon_0 => ("d"phi_"E")/"dt" = 1/epsilon_0 * "dq"/"dt"`
∴ `"I"_"d" = epsilon_0 1/epsilon_0 "dq"/"dt" = "I"`
⇒ Displacement current, `"I"_"d"` = 0.15 A
(c) Yes, Kirchhoff's first law is very much applicable to each plate of capacitor as Id = I.
So current is continuous and constant across each plate.
संबंधित प्रश्न
The charging current for a capacitor is 0.25 A. What is the displacement current across its 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?
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.
Displacement current is given by ______.
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 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 capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by
V = V0sinωt
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.
The charge on a parallel plate capacitor varies as q = q0 cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?
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?
A long straight cable of length `l` is placed symmetrically along z-axis and has radius a(<< l). The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t) = I0 sin (2πνt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is E(s,t) = µ0I0ν cos (2πνt) In `(s/a)hatk`.
- Calculate the displacement current density inside the cable.
- Integrate the displacement current density across the cross-section of the cable to find the total displacement current Id.
- Compare the conduction current I0 with the displacement current `I_0^d`.
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:
