Class 12 Physics Electromagnetic Waves Displacement Current

Displacement Current

• Consider a capacitor and outside the plates of the capacitor there is conduction current IC.
• Area between the plates i.e. inside the capacitor there is displacement current Id.
• Physical behaviour of displacement current is same as that of induction current.
• Difference between Conduction current and Displacement current:-
 Conduction Current Displacement Current It arises due to the fixed charges. It arises due to the change in electric field.
• For Static electric fields:-
• Id=0.
• For time varying electric fields:-
• Id ≠0.
• There can be some scenarios where there will be only conduction current and in some case there will be only displacement current.
• Outside the capacitor there is only conduction current and no displacement current.
• Inside the capacitor there is only displacement current and no conduction current.
• But there can be some scenario where both conduction as well as displacement current is present i.e. I= IC + Id.
• Applying modified Ampere-Maxwell law to calculate magnetic field at the same point of the capacitor considering different amperial loop,the result will be same.

Ampere – Maxwell law: Consequences

Case 1 : Magnetic field is given as

• ∫dl = μ0 Ic
• ∫dl = μ0 Ic / 2πr

Case 2 : Magnetic field is given as

• ∫dl = μ0 Id
• ∫dl = μ0 Id / 2πr

Conclusion: -

1. The value of B is same in both cases.
2. Total current should be the same.
• Time varying electric field generatesmagnetic field given by (Ampere-Maxwell law)
• Consider 1st step up there is electric field between the plates and this electric field is varying with time.
• As a result there is displacement current and this displacement current gives rise to magnetic field.
• Time varying magnetic field generates electric field given by (Faraday-Lenz law)
• Therefore if there is electric field changing with time it generates magnetic field and if there is magnetic field changing with time it generates electric field.
• Electromagnetic waves are based on the above conclusion.

Problem:- Figure shows a capacitor made of two circular plates each ofradius 12 cm, and separated by 5.0 cm. The capacitor is beingcharged by an external source (not shown in the figure). Thecharging current is constant and equal to 0.15A.(a) Calculate the capacitance and the rate of charge of potentialdifference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of thecapacitor? Explain

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 spaceε0, = 8.85 × 10−12 C2 N−1 m−2

(a) Capacitance between the two plates is given by the relation,

C = (ε0A)/ (d)

Where,

A = Area of each plate=πr2

C= (ε0πr2)/ (d)

= (8.85x10-12πx0.12x0.12)/ (0.05)

=8.0032x10-12F

=80.032pF

Charge on each plate, q = CV

Where,

V = Potential difference across the plates

Differentiation on both sides with respect to time (t) gives:

(dq/dt)=C (dV/dt)

Bur (dq/dt) =current (I)

Therefore (dV/dt) = (I/C)

=> (0.15)/ (80.032x10-12) =1.87 ×109 V/s

Therefore, the change in potential difference between the plates is 1.87 ×109 V/s.

(b) The displacement current across the plates is the same as the conduction current.  Hence, the displacement current, id is 0.15 A.

(c) Yes

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

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