Electricity and Magnetism Final Review
Coulomb’s Law Like charges repel, opposite charges attract; net electric charge is conserved. We will label the two kinds of charge as either positive or negative.
Then the force between two point charges is:
F 12 F12
q1
unit vector points from q1 towards q2
q2
+
+
r -
kq 1q 2 r 2 r
the force exerted on charge q2 by the charge q1
F12
r unit vector
r r
rr rr
+
F12
Let the algebra take care of the sign of the force.
Example: Three charges are located at the corners of an equilateral triangle. Find the force on the 7-mC charge. We simply add up the vectors forces. This is easily done by keeping track of the vector components of each force. The force on the 7mC charge due to the 2mC charge:
kq 2q 7 r 2 r 2 14 10 12 C 2 N m 9 9 10 r C2 . 5 2m 2 . 504 cos 60 i sin60 j N . 252 i . 436 j N
F 2,7 F 2,7 F 2,7 F 2,7
The force on the 7mC charge due to the -4mC charge:
F 4,7
12 kq 2q 7 28 10 9 r 9 10 cos 60 i sin60 j r2 . 52
The net force on the 7mC charge is:
. 504 i . 873 j N
F 4,7 . 756 i . 437 j N
The Electric Field Rather than find the force that a distribution of charges exerts on a given (test) charge at a particular point in space, it is convenient to define the ELECTRIC FIELD at this point as the force per unit (positive) charge:
F E q
The Electric Field of a Point Charge
q0
kq E 2 rˆ r
Electric fields originate from positive charges.
The Electric Field of a Point Charge
q0
kq E 2 rˆ r
Electric fields terminate on negative charges.
The Electric Field from a Distribution of Point Charges Again, simply use the superposition principle. To find the Electric Field at a given point in space, simply add up the contributions from each individual charge. Calculate the Electric Field contribution from each one of these charges as if the other ones don’t exist.
kqi E E1 E2 E3 Ei 2 rˆi i i ri
With the charge in an object broken down into particles or small increments dq, we can employ the superposition principle to find the Electric Field at any position. To put it differently we can integrate over the charge increments or charge elements.
k dq dE 2 rˆ r E
dE =
k dq rˆ 2 r
Example: Consider an infinite uniformly charged line with a line charge density l. Find the electric field at a distance y from the line. From the figure we see that the contribution to the x component of the field from charge located symmetrically about the y axis cancel each other. The problem now is to find the contribution to dEy from the charge element dq and integrate over all the charge elements. dE y
kdq kdx sin r2 x2 y2
E y ky Ey
y x2 y2
ky
dx , 3/2 2 2 x y
dx ky 3/2 2 2 x y
x y2
1 1 1 1 . 4 0 y 2 0 y
x2
y2
,
Example: Consider an infinite uniformly charged line with a line charge density l. Find the electric field at a distance y from the line. A few points to be made here: (i) From symmetry the x component vanishes (ii) From symmetry we could have integrated from 0 → ∞ and doubled the result of that integration for Ey. (iii) This integral can also be done with the substitution x=ytanθ → x2+y2=y2/cos2 θ. dx =ydq/ cos2 θ. Try it!
Field vectors that pointing radially outward from an infinite line charge. The magnitude falls off as 1/r.
Example: a uniformly charged ring of radius a and charge Q. Find the electric field on the axis through the ring. Assume that the ring is very thin so that we can regard the charge distribution on it as being a “line charge.” Then the “charge density” (charge per unit length) is l = Q/(2p a). However as we shall see for this case this is superfluous information.
a
P x
dE
k dq k dq cos q r2 x 2 a2 x 2 a2 kxdq E dE x ring ring x 2 a 2 3/2 kxQ kx E dq 3/2 3/2 ring x 2 a 2 x 2 a 2 dE x
Since all of the charge elements in the ring are the same distance from P the integral over dq is simply Q.
k dq rˆ 2 r x
Gauss’ Law The total, net, electric flux through surface R.
Q = sum of all charges enclosed by surface R
Q
E dA
closed surface R
0
0 Permitivity Constant 1
Volt m k = Coulomb' s Constant 9 10 4 p0 C 9
Another example with spherical symmetry: Suppose now that we have a spherically symmetric distribution of charge with charge density r(r) contained within a sphere of radius R. The total amount of charge inside R is Q. R Now surround this distribution 2 Q 4p r r dr with a concentric spherical 0 Gaussian surface with radius r > R. Gauss' Law gives r 2 E d A 4 p r E R sphere of radius r
The same as if Q were a point charge.
Q
0
1 Q This implies that E 4p 0 r 2
Another example with spherical symmetry: Suppose now that we have a spherically symmetric distribution of charge with charge density r(r) contained within a sphere of radius R. The total amount of charge inside R is Q. R Now surround this distribution 2 Q 4p r r dr with a concentric spherical 0 Gaussian surface with radius r < R. Guass’s Law gives
r
R
E dA E E4r 2
4r 2
10
0 r 4r 2dr r
Qenclosed 0
The enclosed charge is the charge inside the sphere of radius r!
Example: “Line Symmetry” Find the electric field outside an infinitely long cylinder of radius R carrying uniform charge per unit length l. Choose a Gaussian surface which is a concentric cylinder.
R
r
E directed everywhere outward, normal to Gaussian surface
electric flux through a lengthl
E dA E cosq dA E dA 2p rl E
l
Apply Guass’s Law 2rlE r l/ 0 E r 1r , r R 2 0
enclosed charge Qenclosed l l
Example: An infinite sheet of charge carries a uniform surface charge density s. Find the electric field arising from the sheet. sheet seen edge-on Pick a Gaussian surface with ends that are parallel to the sheet and have surface area A Enclosed charge Q s A
E
E Apply Gauss's Law implying that E
E dA 2 E A s A /
s 2 0
constant with distance
0
The electric potential from a point charge q r r r
r2
The potential difference between two points at radial distances r1 and r2 , respectively, from the charge q is:
2
1
1
1 E dl 4 p0
1 4 p0
r2 r1 r2 r1
q rˆ dl 2 r q dr 2 r r
q 12 4 p0 r r1 q 1 1 4 p0 r2 r1
r2 q
r1
Note: the potential difference is, of course, independent of the path between the points; so we have chosen the most convenient path to evaluate the integral.
Example: A very long straight wire with radius ra carries a line charge density l.What is the potential difference between the surface of the wire and the ground, a distance rb below?
Treat the wire as an infinitely long, charged rod. Last week we used Gauss’s Law to show that the electric field outside such a rod points radially outward with a magnitude that drops off inversely like the radial distance from the rod’s center line: path increment is l rˆ E dl rˆ dr 2p0 r
DVab
rb
ra
DVab E dl rb
rb
ra
ra
l 2p0
rb ra
l rˆ rˆ dr 2p0 r
dr l rb ln r 2p0 ra
l ra ln 2p0 rb
Example: a charged rod of length ℓ is uniformly charged with a charge Q. (a) What the expression for the potential at a point P located a perpendicular distance r from the center of the rod? The charge element dq is given by dq = l dx = Qdx/ℓ. So the potential is found from the integral:
Vr Vr
rod 2kQ
kdq r2 x2 ln
x
/2
/2
r2
r
x2
kdx r2 x2 /2
0
/2 r 2 2/4 2kQ Vr ln r 2kQ Vr ln /2r 1 2/4r 2
We can generalize this simple example by employing the principle of superposition: if we have many point charges, the total force on any one charge is the sum of the forces from all the others. Therefore, the total energy of the system is the sum of the energies of bringing, sequentially, each charge in from infinity, one at a time. Or, in other words, the total electrostatic energy U is the sum of the energies of all possible pairs of charges:
U
all pairs i,j
1 qi qj 4 0 r ij
i, ji
1 qi qj 4 0 r ij
Example: What is the electrostatic energy of a uniformly charged sphere of radius a and total charge Q? First figure out the uniform volume charge density:
Now imagine that we assemble the charged sphere by bringing in from infinity a succession of infinitesimally thin layers
If q( r ) is the charge of the sphere when it has been built up to radius r, then the in bringing in a thin spherical work done shell with total charge dq and laying it on the surface of the sphere is:
qr r 43 p r 3 dq r 4p r 2 dr
a r dr
Q Q r 4 volume 3 p a 3
1 16p 2 r 2 r 4 dr dU 4p0 3
1 qr dq dU 4p0 r
1 qr dq dU 4p0 r
qr r 43 p r 3 dq r 4p r 2 dr 1 16p 2 r 2 r 4 dr dU 4p0 3
Q qr dq 1 U dU r 4 0 0 2 a 4 1 U r 4dr 4 0 3 0 3 2 3 4a a5 Q2 3 1 U 2 6 5 5 4 0a 4 0 3 a
So, for example, the electrostatic energy of the uniformly charged sphere with total charge Q=0.2 C and radius a = 4 m is:
3
0.2 C 3 Q 5 U 5 4p0 a 4p0 4 m 2
2
2 9 J m 0.2 C 0.69 10 2 C 4 m
5.4 10 7 J
Example: What is the electrostatic energy of a uniformly charged sphere of radius a and total charge Q? Consider integrating the energy density, u = ε₀E²/2, of this system over all space. Q r E r a The electric fields both inside 4 0 a 2 and outside the sphere are: Q 1 Er a 4 0 r 2
Q2 0 2 The energy densities both inside u E r a 2 E r a 24 2 0 and outside the sphere are: Q2 0 2 u E r a E r a 2 24 2 0 Integrating these energy densities over all space leads to
U
0
a
2 Q2 Q2 r 2 4r 2dr 1 4r 2dr 3 Q 2 4 5 4 0a a 24 0 r 24 2 0 a 4
r2 a6 1 r4
Capacitance As a consequence of the superposition principle, and as we have seen in numerous examples, the potential difference V between conducting bodies is proportional to the charge Q carried by them. It is customary to write this proportionality in what at first seems like a backward way:
QC V
potential difference between bodies
capacitance (sometimes “capacity”)
+Q
Q -Q
two conducting bodies
Can we talk about the capacitance of an isolated conductor with charge Q? Yes, however in this case V is the potential difference relative to infinity.
Example: A long coaxial cable consists of an inner cylindrical conductor of radius a and an outer cylindrical conductor of radius b. Find the electrostatic energy and capacitance for a cable of length ℓ the conductors carry equal and opposite linear charge densities of ±λ. From Guass’s law we know that the field between the conductors is given by
Er
The energy density is
Integrating this energy density over a length ℓ
1 2 0 r
2 2 1 1 1 1 2 u 0E r 0 2 2 2 2 2 4 0 r 8 2 0 r 2
2 U udV 8 2 0
a
b
2rdr 2 ln b a 4 0 r2
Example: Find the capacitance, c, per unit length for this same coaxial cable. Again from Guass’s law we know that the field between the conductors is given by
Er
1 2 0 r
Va Vb Edr a
The potential difference between the conductors is
Since the charge per unit length is λ
a
b
dr 2 0r b Va Vb V ln b a 2 0
2 0 c V lnb/a
The total capacitance is proportional the length of the cable, or C=cL.
Resistance and Ohm’s Law a segment of wire with potential drop V and length l and area A and conductivity s If the wire is uniform, there will be a uniform electric field and, hence, a resulting uniform current density
V E j E
The total current is the surface integral of j through the end cap
I
Using the above relation between potential and electric field magnitude
Ohm’s Law (macroscopic version)
j dA jA EA cap
I VA V
V RI
I A
R A A
Electric Power If a potential difference V is applied across a resistor and a current I flows, the we can conclude that the power dissipated in heat is:
P VI The potential drop across the resistor is the energy gained per unit charge as that charge falls through the potential difference; the current is the rate at which charge falls through this potential difference. Using Ohm’s law in this expression for the power, we find
2 V P I 2R R
The dimensions of power are, of course, Watts, or Joules per second.
V IR
Adding resistances in parallel and in series: Parallel A
R1
Series
in both cases, potential drop from A to B is VAB
R2
B Parallel Case: each resistor has the same potential across it:
VAB = I1R1 VAB I2R2 VAB VAB 1 1 I I1 I2 VAB R1 R2 R1 R2
1 1 1 R R1 R2
A
R1
R2 B Series Case: each resistor has the same current through it:
VAB V1 V2 R1I R2 I RI
R R1 R2
Lorentz Force Law We can define the magnetic field B (a vector quantity) at a point by the vector force Fmag at that point experienced by a point particle with charge q and velocity v:
Fmag q v B If there is also an electric field E at this point, then in addition to the above magnetic force, there will be an electric force Felec=qE and the total force Ftot on the charge will be
Ftot = Felec Fmag q E q v B qE v B This is the most general form of the Lorentz Force Law.
In this case, we have uniform circular motion with the centripetal acceleration supplied by the magnetic force:
Fmag qvB sinq qvB sin90 qvB
m v2 Fcentrip r
Fmag Fcentrip mv qvB r
v
Fmag
where r is the radius of the circular path
r
Since vT = 2pr we find:
2
mv r qB vT 2mv T 2m qB qB qB f qB/m 2m Cylotron Frequency
The Magnetic Force on a Current What about the case where the current does not travel in a straight line? That is the wire follows a curved path. In this case we define a short section of the wire containing a charge dq. The force on this section of wire due to a magnetic field is : dq dF dqv d B dq dl B dl B dt dt dF Idl B
current I dl
B-field
Now the vector dl is differential length of the wire pointing in the direction of the current. The total force is just the sum the forces along the conductor.
F
dF Idl B
We note that all quantities, (vector) magnetic field, (vector) infinitesimal line segment, and current could depend on position and must therefore, in general, be under the integral sign.
We had the following relation, which is called the Law of Biot & Savart. You can see that it is a kind of “Coulomb’s Law” for magnetic fields. Like Coulomb’s Law, it is valid only when fields are not changing in time
Br1 0 4
Let us say that we have a thin uniform wire with cross sectional area A and arbitrary shape and carrying current I.
jr2 r 12 3 d r2 2 r1 r2
Contribution from wire increment
jr2 d 3r2 = j dVol jA dlr2 I dlr2
0 Idlr2 r 12 dBr1 4 r1 r2 2
A
Summing over the entire wire
r 12
j
r2
dlr2
r1
0 Br1 4
Idlr2 r 12 r1 r2 2
Example: Find the magnetic field along the symmetry axis at a distance x from the center of a ring carrying a current I and radius a as shown. From the Biot-Savart law the field from an incremental current element is
0 Id r dB 4 r2
The integration is now straightforward (trivial actually):
Ia Bz 0 4
0
2
d 0Ia 2 z cos z z2 a2 2z 2 a 2 3/2
Since a current loop behaves as a magnetic dipole we find for x>>a B 0 3 where IA Ia 2 2 z This expression holds at large distances independent of the shape of the current loop. Off axis calculations confirm that the angular dependence is that of a dipole as well.
Ampere’s Law The figure below shows a path ADCBA for the integration of Bdℓ . In this case our line integral still remains:
B d 0I The scalar product along AD and CB vanishes and the integration along DC only depends on the angular separation which is identical to the angular separation for the path AC. In fact the integral
B d is independent of the path taken around the line current! This leads us to Ampere’s Law, which only holds for steady state currents:
B d 0I encircled
Example: Magnetic field inside a wire. A long straight wire of radius R carries a current I with uniform current density j. Find the magnetic field as a function of distance from the center of the wire. All of the cylindrical symmetries are still present. Hence it is straightforward to make use of Ampere’s law.
B d 0I encircled 0 j dA Since the current density is uniform, j = I/(pR2). Performing the integrals above:
B 2r 0 I 2 r 2 R 0I r B rR 2 R 2
Outside the wire the field is that which we found on the previous slide:
Example: Magnetic field due to a sheet of current. An infinite flat sheet carries current with a uniform current density out of the page as shown. Find the magnetic field as a function of distance from the center of the sheet . Again from symmetries it is straightforward to apply Ampere’s law. Consider the amperian loop as shown.
B d 0I encircled 0 j dA Assume a thickness w for the sheet. Since the current density is uniform we define, j = I/(wℓ)=Js/w, so that Js=I/ℓ is the current per width of the sheet. Performing the integrals above:
2B 0J s B 1 0J s 2 The magnetic field is independent of the distance from the sheet. Also it points in opposing directions on opposite sides of the sheet. (remember the RHR) Note the analogy with the electric field of an infinite charged plane. These sheets of current are good approximations to ribbons of current e.g. wide flat conductors. They also form in nature in conducting plasmas.
Solenoids and Toroids The figure is the cross section of a tightly wound current carrying coil. The resulting magnetic field is closely approximated by two conducting sheets. As such we would expect the magnetic field to be a superposition of two sheets:
B 0J s Consistent with superposition, the field outside the coil vanishes. Consider the amperian loop as shown in the figure. Applying Ampere’s law:
B d B 0nI where n # turns per unit length B 0nI For a current sheet the total current passing through the amperian loop satisfies:
J s I tot nI J s nI
So we have a consistent result.
Faraday’s Law (the third of Maxwell’s equations)
dB EMFinduced E dl dt closed loop
This minus sign constitutes “Lenz’s Law.”
The physical meaning of this minus sign (Lenz’s Law): direction of induced EMF and associated current (if any). A current induced by a changing magnetic flux is in the direction which produces a magnetic field which opposes the sense of change of the magnetic flux. Energy conservation demands this minus sign: power dissipated by a current from an induced EMF must be compensated by a positive rate of work done in changing the magnetic flux.
Example: Consider a pair of conducting rails a distance ℓ apart in a uniform magnetic field B, which points into the screen. A resistance R is connected across the rails, and a conducting bar of negligible resistance is being pulled along the rails with velocity v to the right. (a) What is the current in the circuit? (b) What is the direction of the current through the resistor? (c) At what rate must work be done by the agent pulling the bar? (a) First we make use of Faraday’s law: The emf via Faraday’s law is:
d B dt E d Bx B dx dt dt E Bv E E dl
So the magnitude of the current is simply:
|I| |E|/R Bv/R
(b) From Lenz’s law, the current must flow in the counter clockwise direction. The field induced by the emf is in the opposite direction as the applied field. (c) The rate of work done is simply:
P EI E 2/R Bv 2/R
Self Inductance If there are currents flowing (and possibly varying in time) in both circuits simultaneously, then the magnetic flux through either of the circuits individually will be the sum of the separate magnetic fluxes from the currents flowing in both circuits. The induced EMF in one of the circuits therefore will depend on the rate of change of current in both circuits. For example, we can write the induced EMF in circuit 2 as
E 2 M dI 1 L dI 2 dt dt
where L is the self inductance of circuit 2.
Let us consider circuit 2 as completely isolated (e.g., circuit 1 doesn’t exist). Any circuit has an interesting property: its own magnetic field (stemming from its own current) produces a magnetic flux through the circuit which opposes changes in the current. This is self inductance, defined as the ratio of the magnetic flux through the circuit to the current
B L I
Example: Problem 11, Chapter 32: What is the self inductance of a solenoid 50 cm long and 4.0 cm in diameter that contains N = 1000 turns of wire? Solution: the magnetic field in the solenoid is taken to be uniform with a magnitude, as we have seen before, that depends on the number of turns per unit length n B m nI 0
In this case because the field is uniform and normal to the plane of each turn, the magnetic flux through each turn is just the product of magnetic field magnitude and the cross sectional area of the solenoid each turn 2
BA Bp r
B
The magnetic flux through the whole solenoid is then: each turn 2 2 total N NB p r N( m nI) p r B B 0
N 2m0 I p r 2 l
where n N /l The self inductance is then
L
total B
I
m0 N 2 p r 2 l
4p 10
7
J C V C Vs A Vs m m sm m
Vs A
N/A 1000 p 0.02 m m 3 3.16 10 H A 0.5 m 2
2
2
Summary and Impedance/Reactance The (complex) voltage across the circuit elements are all of the form:
V Z I where Z R R, Z C 1/iC, Z L iL The quantity Z is called the impedance of the circuit element and plays the role of resistance. However for an inductor and capacitor it is frequency dependent and includes a phase shift. From the form of the impedance we see that the phase of I leads the phase of V by p/2 in a capacitor and lags by p/2 in an inductor. The magnitude of the impedance is defined as the reactance so that XC = 1/wC and XL = wL where the symbol X is defined as the reactance. Taking the ratio of the peak voltage to the peak current involves only the magnitude of the impedance and we find
V p XI p V p /I p X
Driven RLC Circuit It is of interest here to note that in the frequency regime we can write
V p Z tot I 0 Z R Z L Z C I 0 V p R iL 1/C I 0
V V p cos t
Here we have the same result as obtained solving the differential equation. Hence the impedance approach precludes solving this differential equation! Now we find
I0 |Z|
Vp V p e i |Z| R iL 1/C R 2 L 1/C 2 , tan
L 1/C R
Driven RLC Circuit Hence the real part of the current is
It Re I
V V p cos t
Re I 0e it
I 0 Ree it
It I 0 cost , I 0 V p /|Z| What about the voltages across the elements in the circuit?
These are simply the real part of the impedance times the complex current. For example the voltage across the resistor is
V R Z R I R I RI 0e it V R RI 0 cost R V p cost |Z|
VC ZC I and the capacitor:
1 I I 0 e it/2 iC C
Vp cost /2 C|Z| Vp VC sint C|Z| VC
How to think about electricity and magnetism . . . The physical things that are measured in a lab have to do with the forces on (and consequent accelerations of) charges, i.e., the flow of momentum. The Lorentz Force Law gives these and, in so doing, serves to define the electric and magnetic fields.
F qE q v B
Having so defined the electric and magnetic fields, there are four relations which relate these to charges and currents. These are Maxwell’s Equations: TRUE ALWAYS
E dA
Qenclosed
0
closed surface
B dA 0
The magnetic flux through a closed surface is zero.
closed surface
E dl closed loop
B dl closed loop
The electric flux through a closed surface is equal to the charge enclosed (divided by a constant).
B dA t surface enclosed by loop
The line integral of the electric field around a closed path is equal to minus the rate of change of magnetic flux through any surface which has the path as its boundary.
1 E dA m0 Ithrough 2 c t surface surface enclosed by loop
The line integral of the magnetic field around a closed path is equal to the rate of change of electric flux plus the net current flowing through any surface which has the path as its boundary.
Differential Form of Maxwell’s Equations The more common form of Maxwell’s equations involves differential vector operators, the divergence and the curl.
V FdV F dA
The divergence theorem or Guass’s theorem states:
Consider Guass’s law:
S
q
E dA enc 0
Since the integration is over an arbitrary volume this reduces to:
V EdV
1 0
V dV
E 0
An analogous relation holds for the magnetic field, ∇⋅B = 0. However in this case the magnetic charge density is always zero. These two scalar relations make up two of the differential forms of Maxwell’s equations.
Differential Form of Maxwell’s Equations The more common form of Maxwell’s equations involves differential vector operators, the divergence and the curl.
S F dA
Stoke’s theorem states:
Consider Faraday’s law
F dl loop
E dl B t
loop
Since the integration is over an arbitrary surface area this reduces to
Application of Stokes theorem to Ampere’s law leads to
S E dA t
B dA S
E B t
B 0j 0 0 E t
These four differential forms for Maxwell’s equations plus the Lorentz force law are all that is required to describe all of classical electromagnetism!
Thatâ&#x20AC;&#x2122;s All Folks! This completes Physics 2B. The material was challenging but interesting. I hope you enjoyed at least some of it or at least learned some of it! There will be a review session Thursday that should be helpful in preparing for the final! Good Luck to Everyone!