AP Physics C - Maxwell’s Equations and Electromagnetic Waves

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/ε₀

[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.dA over any closed surface is 1/ε₀ 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].

4. ∫_{closed path}B.dℓ = μ₀i_c + μ₀ε₀dФ_E/dt

[This is Ampere’s circuital law with Maxwell’s modification. The firest term, μ₀i_c on the right hand side contains the conduction current i_c. The second term, μ₀ε₀(dФ_E/dt) was added by Maxwell to incorporate the displacement current i_d = ε₀(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/ε₀ becomes ∫_v div Edv = (1/ε₀) ∫_v ρdv, on applying Gauss divergence theorem to the left hand side of the equation and by putting Q = ∫_v ρdv where ρ is the volume charge density. The volume integration is done over the volume ‘v’ enclosed by the closed surface. Therefore,

div E = ρ/ε₀

This can be written also as

div D = ρ

where D is the electric displacement vector which in free space is given by D = ε₀E.

(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 = μ₀i_c + μ₀ε₀ (dФ_E/dt) becomes ∫_s curl B.dA = ∫_s μ₀J. dA + ∫_s μ₀ε₀(d/dt) E. dA, on applying Stokes theorem to the left hand side of the equation and by putting i_c = ∫_s J . dA and Ф_E= ∫_s E. dA. Therefore,

curl B = μ₀J + μ₀ε₀(dE/dt) = μ₀ (J + dD/dt) since D = ε₀E 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 = ε₀A/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 = (ε₀A/d)×d×dE/dt = ε₀A dE/dt = ε₀dФ_E/dt, on substituting for the electric flux Ф_E = AE.

The above steps show that the quantity ε₀dФ_E/dt indeed 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×B points 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/√(μ₀ ε₀)

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

v = 1/√(μ ε)

Since μ = μ₀ μ_r and ε = ε₀ ε_r where μ_r and ε_r are respectively the relative permeability and the relative permittivity 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 = ½ ε₀E_rms² 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_rms²/μ₀) where B_rms is the root mean square value of the magnetic field.

U_E and U_B are equal since B_rms = E_rms/c = E_rms√(μ₀ ε₀)

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

U = U_E + U_B = ½ ε₀E_rms² + ½ (B_rms²/μ₀)

Since U_E = U_B, we have

U = ε₀E_rms² = B_rms²/μ₀

In the case sinusoidally oscillating electric and magnetic fields E_rms = E_m/√2 and B_rms = B_m/√2 so that the energy density can be written as

U = ε₀E_m²/2 = B_m²/2μ₀

(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 = ε₀c E_m²/2 = (E_m²/2)√(ε₀/μ₀) since c =1/√(μ₀ ε₀)

This can also be written as

I = E_m²/2cμ₀

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

S = E×B/μ₀

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/μ₀ which is equal to E²/cμ₀. 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.

AP Physics C- Additional Multiple Choice Practice Questions on Rotational Motion

In the post dated 20^th January 2008, equations to be remembered in circular motion and rotation were discussed. Subsequently some multiple choice practice questions on circular motion and rotation were discussed in the posts dated 24^th January 2008, 26^th January 2008 and 9^th May 2009. Free response practice questions in this section were discussed in the posts dated 23^rd January 2008 and March 7^th 2009. You can access all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.

Today we will discuss a few more multiple choice questions in this section:

(1) A solid sphere of radius r is released (from rest) from the top inner edge (position P in fig.) of a hemispherical bowl. The sphere and the bowl have smooth surfaces. What will be the angular velocity of the sphere about the centre O of the hemispherical bowl when the sphere reaches the bottom B of the bowl?

(a) [2g(R – r)]^1/2

(b) [2g/(R – r)]^1/2

(c) [10g/ 7(R – r)]^1/2

(d) [10g/ 7(R – r)]

(e) [2g/5(R – r)]^1/2

The important thing you need to remember is that there cannot be any rolling in the absence of friction. The solid sphere will simply slide along the inner surface of the hemispherical bowl. The problem is therefore simpler than some of you might have imagined.

The centre of gravity of the sphere has come down through a distance R – r on reaching the bottom B of the hemispherical bowl. Consequently, the loss in the gravitational potential energy of the sphere is mg(R – r) where m is its mass. The sphere gains an equal amount of kinetic energy so that we have

½ mv² = mg(R – r) where v is the velocity of the sphere at the bottom B of the bowl. This gives v = [2g(R – r)]^1/2.

[Normally you will remember the speed v = √(2gh) in the case of a body falling freely from a height h (= R – r here) and you can skip the above steps while working out multiple choice questions].

Angular velocity ω of the sphere about the centre O of the hemispherical bowl is given by

ω = v/(R – r) = [2g(R – r)]^1/2/(R – r) = [2g/(R – r)]^1/2

(2) If the surfaces of the sphere and the bowl in the above question are rough and the sphere rolls down without slipping, what will be the angular velocity of the sphere about the centre O of the hemispherical bowl when the sphere reaches the bottom B of the bowl?

(a) [2g(R – r)]^1/2

(b) [7g/ 5(R – r)]^1/2

(c) [10g/ 7(R – r)]^1/2

(d) [10g/ 7(R – r)]

(e) [2g/5(R – r)]^1/2

The loss of potential energy of the sphere is mg(R – r) as in the above question. But the kinetic energy in this case has two parts: translational K. E. and rotational K. E. Therefore, we have

½ mv² + ½ I ω_s² = mg(R – r) where I is the moment of inertia of the solid sphere [I = (2/5)mr²] and ω_s is the spin angular velocity of the sphere (about its own axis). Since ω_s = v/r the above equation becomes

½ mv² + ½ ×(2/5)mr²×(v/r) ² = mg(R – r)

This gives 7v²/10 = g(R – r) so that v = [10g(R – r)/ 7]^1/2

The angular velocity of the sphere about the centre O of the hemispherical bowl is given by

ω = v/(R – r) = [10g(R – r)/ 7]^1/2/(R – r) = [10g/ 7(R – r)]^1/2

(3) A circular disc of mass M and radius R is at rest at the top of an incline of height H (Fig.). On releasing, the disc rolls down the incline without slipping. What is the angular momentum of the disc about its centre of mass when it reaches the bottom of the incline?

(a) 2MR √(gH/3)

(b) (M/R) √(gH/3)

(c) √(2MgH/3)

(d) M √(gH/3R)

(e) MR √(gH/3)

You have to first find out the spin angular velocity ω using appropriate expression for the moment of inertia I of the disc (I = ½ MR²). Additionally, you are required to calculate the angular momentum L = Iω.

The linear velocity v of the disc at the bottom of the incline is given by

½ Mv² + ½ I ω² = MgH

But v = ωR so that

½ Mω²R² + ½ ×½ MR²ω² = MgH

Or, ¾ ω²R² = gH from which ω =√(4gH/3R²)

Angular momentum L of the disc about its centre of mass is given by

L = Iω = ½ MR²×√(4gH/3R²) = MR√(gH/3)

Answer to AP Physics C Free Response Practice Question on Electric Field & Potential

In the post dated 2^nd August 2009, the following free-response question for practice was given to you:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

As promised, I give below a model answer for the above question:

(a) The electric field at the centre of the ring is zero.

This follows from the definition of the electric field: Electric field at any point is the force per unit positive test charge placed at the point. The test charge placed at the centre of the ring will be repelled equally by the uniformly distributed positive charges on the conducting ring so that the net force on the test charge will be zero.

[Note that the charge Q placed at the position A on the conducting ring will be uniformly distributed immediately throughout the ring].

(b):

(i) The charges on the ring are at the same distance R from the centre of the ring and hence the potential at the centre of the ring is Q/4πε₀R.

(ii) If the charge placed on the ring is negative, the potential will be negative. The potential in this case will be – Q/4πε₀R.

(c) Consider a small quantity (dQ) of charge at a point such as A on the ring. The electric potential dV at P due to this elemental charge is given by

dV = dQ/4πε₀r.

The total potential (V) at P due to all the charges on the ring is given by

V = ∫dV = ∫dQ/4πε₀r = Q/4πε₀r (since all charge elements are at the same distance r from the point P).

(d) The electric field at P due to an element dQ of charge at a point such as A is directed along AP. This field can be resolved into two components: one component along the axis of the ring (axial component) and the other component perpendicular to the axis of the ring (normal component). For every elemental charge at any given point on the ring, there is an equal elemental charge situated diametrically opposite to it. The diametrically opposite element will produce an equal axial component of field in the same direction. But the normal component due to the diametrically opposite elemental charge will be equal in magnitude but opposite in direction. Therefore, all axial components get added where as all normal components get canceled.

The electric field due to the charged conducting ring is therefore directed along the axis of the ring.

The electric field at the point P is given by

E = –grad V.

Since we know that the electric field is along the axis of the ring which is along the x-axis as given in the question, we have

E = –∂V/∂x = –∂/_∂x (Q/4πε₀r) = –∂/_∂x [Q/4πε₀√(R²+x²)]

Or, E = (1/4πε₀) [Qx/(R²+x²)^3/2]

Since the charge on the ring is positive, the field E is along the positive x-direction.

When the electric field is maximum, dE/dx = 0

Therefore, (1/4πε₀)[{(–3/2)(R²+x²)^–5/2 ×2x ×x} + (R²+x²)^–3/2] = 0

Or, 3x² (R²+x²)^–1 = 1

Or, 3x² = R²+x² from which x = R/√2.

(e) The nature of variation of the electric field along the axis of the ring is shown in the adjoining figure in which O represents the centre of the ring. The field has maximum magnitude at points distant R/√2 from the centre and these are located symmetrically on either side of the centre. Positive direction of the field is the positive x-direction even though it is represented by positive y-plot in the figure. The negative direction of the field is the negative x-direction and is represented by negative y-plot in the figure.

You can access all posts related to electrostatics (including multiple choice practice questions) on this site by clicking on the label ‘electrostatics’ below this post.

Electrostatics- AP Physics C Free Response Practice Question on Electric Field & Potential

Essential points to be remembered in respect of electric field and potential were discussed on this blog on 16^th June 2008. You can access that post by clicking here.

All posts related to electrostatics can be accessed by clicking on the label ‘electrostatics’ below this post.

Today I’ll give you a free response practice question involving electric field and potential:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

This question carries 15 points and you have 15 minutes for answering it. Try to answer this question. I’ll be back soon with a model answer for your benefit.