As promised in the post dated 26^{th} August 2009, I give below a few typical multiple choice practice questions for AP Physics C students:

(1) The displacement current in the dielectric between the plates of a 0.1 μF capacitor is 1 mA. The potential difference between the plates of this capacitor must be changing at the rate of

(a) 10^{4} volt/second

(b) 10^{3} volt/second

(c) 10^{2} volt/second

(d) 10 volt/second

(e) 1 volt/second

This is a simple question contrary to the impression of some of you at the first glance! Whether it is conduction current or displacement current, we have the usual expression for electric current (i) as

i = dQ/dt

Since the charge Q = CV, we have i = CdV/dt

This gives dV/dt = i/C = 1 mA /0.1μF = 10^{–3} ampere^{ }/10^{–7} farad = 10^{4} volt/second [Option (a)].

(2) A 10 μF electrolytic capacitor connected to a 6 V power supply is fully charged. The displacement current flowing through the dielectric is

(a) 100 mA

(b) 10 mA

(c) 1mA

(d) 0.1 mA

(e) zero

The correct option is (e) since there is no charging current in the case of the fully charged capacitor.

(3) The electric and magnetic fields in a radio wave propagating along the z-direction are given respectively by

E_{x}= E_{0} sin (kz–ωt ) and

B_{y}= B_{0} sin (kz–ωt )

where E_{0} and B_{0} are the amplitudes, k is the magnitude of the wave vector (propagation vector) and ω is the angular frequency. How is E_{0} related to B_{0}?

(a) E_{0}=B_{0}kω

(b) E_{0}= B_{0}k/ω

(c) E_{0 }= B_{0}√(kω)

(d) E_{0}= B_{0}

(e)E_{0 }= B_{0}ω/k

The electric and magnetic fields in the wave are varying simple harmonically and the speed of the wave is given by c = ω/k. [Note that the speed of a wave propagating in the z-direction is given by the ratio of coefficient of t to the coefficient of z. [See the post dated 9^{th} April 2009 ‘AP Physics B- Wave Motion (including sound)- Points to be Noted’ which you can access by clicking on the label ‘waves’ below this post.

Since E = Bc we have E_{0 }= B_{0}c = B_{0}ω/k

(4) The charge on the plates of a capacitor of plate area 0.1 m^{2} is 0.05 coulomb and is decreasing at the rate of 0.5 coulomb per second at an instant t during its discharge. What is the displacement current at the instant t?

(a) 0.005 A

(b) 0.05 A

(c) 0.5 A

(d) 5 A

(e) Data insufficient for arriving at an answer

Data is more than sufficient. The plate area given in the question is not required and it serves the purpose of a distraction. The displacement current is the time rate of change of charge (dQ/dt) on the plates and is equal to 0.5 ampere [Option (c)].

[If the capacitor is an ideal parallel plate capacitor and you are required to find the magnitude of the displacement current density, the answer is 0.5 A/0.1 m^{2} = 5 Am^{–2}]

(5) At a particular point in space and time, the magnetic field in a plane electromagnetic wave traveling in free space along the positive x-direction is given by B = 3.1×10^{–8}k T (where i, j, k are unit vectors along the x, y, z directions as usual). The electric field at the same point at the same instant is

(a) – 1.033 j Vm^{–1}

(b) 9.3 i Vm^{–1}

(c) – 9.3 j Vm^{–1}

(d) 1.033 k Vm^{–1}

(e) 9.3 j Vm^{–1}

The magnitude of the electric field is given by E = Bc = 3.1×10^{–8 }×3×10^{8} Vm^{–1} = 9.3 Vm^{–1}. Since the wave is proceeding along the positive x-direction and the magnetic field vector is along the positive z-direction (as indicated by the unit vector k), the electric field must be along the positive y-direction. [Note that the cross product vector E×B should point along the direction of propagation].

Therefore, E = 9.3 j Vm^{–1}.

(6) A microwave signal traveling through the earth’s atmosphere has an electric field of amplitude 0.02 Vm^{–1}. The intensity (power flow per unit area) of this signal is (Permittivity of free space, ε_{0 }= 8.85×10^{–12} C^{2 }N^{–1 }m^{–2})

(a) 5.31×10^{–2} Wm^{–2}

(b) 5.31×10^{–3} Wm^{–2}

(c) 5.31×10^{–6} Wm^{–2}

(d) 5.31×10^{–7} Wm^{–2}

(e) 5.31×10^{–8} Wm^{–2}

The intensity (I) which is the power flow through unit area (with the plane of the area perpendicular to the direction of propagation) is given by

I =ε_{0 }c_{ }E_{m}^{2}/2 where c is the speed of the signal and E_{m} is the amplitude of the electric field.

Therefore, I =(8.85×10^{–12} ×3×10^{8}×0.0004) / 2 = 5.31×10^{–7} Wm^{–2}

This post is meant for AP Physics C aspirants who are required to have some idea about Maxwell’s equations and their consequence.

Maxwell’s equations are basically the mathematical statements of

(i) Gauss’s law in electricity

(ii) Gauss’s law in magnetism

(iii) Faraday’s law of electromagnetic induction and

(iv) Ampere-Maxwell law.

The last one is the well known Ampere’s law with Maxwell’s modification for incorporating displacement current (which can flow even through empty space), in addition to conduction current (which flow through conductors). Here are Maxwell’s equations:

1. ∫_{closed surface}E.dA = Q/ε_{0}

[This is Gauss’s law in electricity which states that the flux of the electric field E through any closed surface, that is, the surface integral of E.dAover any closed surface is 1/ε_{0} times the total charge Q enclosed by the surface. Note that E is the electric field vector present at an elemental area vector dA of the closed surface].

2. ∫_{closed surface}B.dA = 0

[This is Gauss’s law in magnetism which states that the magnetic flux through a closed surface is zero].

3. ∫_{closed path}E.dℓ = – dФ_{B}/dt

[This is Faraday’s law of electromagnetic induction which states that the induced emf in a circuit is equal to the time rate of change of magnetic flux Ф_{B}. The negative sign is because of Lenz’s law which states that the induced emf opposes the change of the magnetic flux].

[This is Ampere’s circuital law with Maxwell’s modification. The firest term, μ_{0}i_{c} on the right hand side contains the conduction current i_{c}. The second term, μ_{0}ε_{0}(dФ_{E}/dt) was added by Maxwell to incorporate the displacement current i_{d} = ε_{0}(dФ_{E}/dt). Note that the displacement current is produced because of the time rate of change of the electric field].

Maxwell’s equations given above are in the integral form. The differential form of Maxwell’s equations can be easily obtained by applying Gauss’s divergence theorem and Stokes theorem. Thus we have the following equations:

(i) ∫_{closed surface}E.dA = Q/ε_{0} becomes ∫_{v }div Edv = (1/ε_{0}) ∫_{v}ρdv, on applying Gauss divergence theorem to the left hand side of the equation and by putting Q = ∫_{v}ρdvwhere ρ is the volume charge density. The volume integration is done over the volume ‘v’ enclosed by the closed surface. Therefore,

div E = ρ/ε_{0}

This can be written also as

div D = ρ

where D is the electric displacement vector which in free space is given by D = ε_{0}E.

(ii) The second equation,∫_{closed surface}B.dA = 0 similarly becomes

div B = 0

(iii) The 3^{rd} equation, ∫_{closed path} E.dℓ = – dФ_{B}/dt becomes ∫_{s} curl E.dA = ∫_{s} (d/dt) B. dA, on applying Stokes theorem to the left hand side of the equation and by putting the magnetic flux Ф_{B} = ∫_{s}B. dA. The surface integration is performed over the entire area A enclosed by the closed path. Therefore,

curl E = – dB/dt

(iv) The 4^{th} equation, ∫_{closed path}B.dℓ = μ_{0}i_{c} + μ_{0}ε_{0 }(dФ_{E}/dt) becomes ∫_{s} curl B.dA = ∫_{s }μ_{0}J. dA + ∫_{s} μ_{0}ε_{0}(d/dt) E. dA, on applying Stokes theorem to the left hand side of the equation and by puttingi_{c} = ∫_{s} J . dA andФ_{E}= ∫_{s}E. dA. Therefore,

curl B = μ_{0}J + μ_{0}ε_{0}(dE/dt) = μ_{0 }(J + dD/dt) since D = ε_{0}E for free space.

Displacement current:

The concept of the displacement current was introduced by Maxwell from his understanding that all electric currents must be closed. For instance, in the charging of a capacitor, a conduction current i_{c} flows in the wires connecting the capacitor to the charging battery and an equal (total) displacement current flows through the dielectric (or free space) in between the capacitor plates. The conduction current in the connecting wire and the displacement current in the space between the plates of the capacitor make a closed current circuit.

The displacement current (i) as well as the conduction current is given, as usual, by

i = dQ/dt

But Q = CV where C is the capacitance and V is the voltage across the capacitor. Therefore dQ = CdV so that the displacement current i = CdV/dt.

In the simple case of a parallel plate capacitor with air or free space between the plates, C = ε_{0}A/d where A is the area of the plates and d is the separation between the plates. Further, V = Ed where E is the electric field between the plates. Therefore, displacement current i =(ε_{0}A/d)×d×dE/dt = ε_{0}A dE/dt = ε_{0}dФ_{E}/dt, onsubstituting for the electric flux Ф_{E} = AE.

The above steps show that the quantity ε_{0}dФ_{E}/dtindeed represents the displacement current.

Electromagnetic Waves:

Electromagnetic waves are produced by accelerated charges. An oscillating electric charge produces an oscillating electric field in space, which produces an oscillating magnetic field. But an oscillating magnetic field is a source of oscillating electric field. Therefore an oscillating electric charge can produce oscillating electric and magnetic fields which regenerate each other and an electromagnetic wave propagates through the space.

The electric field and the magnetic field in an electromagnetic wave are perpendicular to each other. (Note that in charging a capacitor, the electric field in the space between the capacitor plates is directed perpendicular to the plates where as the magnetic field produced by the displacement current flowing through the space between the capacitor plates is along circles around the electric field lines).

In the case of a plane electromagnetic wave propagating along the positive z-direction, the electric field is along the positive x-direction and the magnetic field is along the positive y-direction.

Remember:

(i) The vectors E and B in an electromagnetic wave are perpendicular to each other and so oriented that the vector product (cross product) E×Bpoints in the direction of propagation of the wave.

As an example of the application of this rule, suppose the electric field vector is along the negative z-direction and the magnetic field vector is along the positive x-direction. This wave has to be propagating along the negative y-direction (Fig.).

(ii) The magnitudes (E and B) of the electric field and the magnetic field in an electromagnetic wave are related as

B = E/c

where c is the speed of electromagnetic waves.

(iii) The speed (c) of electromagnetic waves in free space is given by

c = 1/√(μ_{0}ε_{0})

(iii) The speed (v) of electromagnetic waves in a medium of permittivity ε and permeability μ is given by

v = 1/√(με)

Since μ = μ_{0} μ_{r}and ε = ε_{0} ε_{r} where μ_{r} and ε_{r}_{ }are respectively the relative permeability andthe relativepermittivity of the medium, we have

v = c/√(μ_{r}ε_{r})

(iv) The energy density (energy per unit volume) in the region of space through which an electromagnetic wave propagates is due to the electric and magnetic fields associated with the wave. The energy density (U_{E}) due to the electric field is given by

U_{E} = ½ε_{0}E_{rms}^{2} where E_{rms} is the root mean square value of the electric field.[We use the root mean square value since the field is oscillating].

Similarly the energy density (U_{B}) due to the magnetic field is given by

U_{B} = ½(B_{rms}^{2}/μ_{0})where B_{rms} is the root mean square value of the magnetic field.

U_{E} and U_{B} are equalsince B_{rms} = E_{rms}/c = E_{rms}√(μ_{0}ε_{0})

The energy density (U) due to the electromagnetic wave is the sum of the above energy densitiesand is given by

U = U_{E} + U_{B} = ½ε_{0}E_{rms}^{2} + ½(B_{rms}^{2}/μ_{0})

Since U_{E} = U_{B}, we have

U = ε_{0}E_{rms}^{2} = B_{rms}^{2}/μ_{0}

In the case sinusoidally oscillating electric and magnetic fields E_{rms} = E_{m}/√2 and B_{rms} = B_{m}/√2 so that the energydensitycan be written as

U =ε_{0}E_{m}^{2}/2 = B_{m}^{2}/2μ_{0}

(v) The intensity (I) of the electromagnetic wave, which is the power flow through unit area, that is, the energy flowing per second through unit area (the plane of the area held perpendicular to the direction of propagation of the wave) is given by

I = Uc = ε_{0}c_{ }E_{m}^{2}/2 = (E_{m}^{2}/2)√(ε_{0}/μ_{0}) since c =1/√(μ_{0}ε_{0})

This can also be written as

I = E_{m}^{2}/2cμ_{0}

(vi) The power flow through unit area is described by Poynting vector S given by

S = E×B/μ_{0}

Poynting vector S is directed along the direction of propagation of the wave. Since E and B are perpendicular to each other, Poynting vector has magnitude EB/μ_{0} which is equal to E^{2}/cμ_{0}. Note that this quantity represents the instantaneous power flow through unit area.

In the next post we will discuss questions in this section. Meanwhile see these posts at physicsplus.