Section 2-1 - Electric Charges, Forces, & Fields

Interactions between Materials & Franklin's One Fluid Model
A Twentieth Century Explanation
Properties of Charge
Coulomb's Law
The Electric Field
Gauss's Law
Metals
Dipoles
Correlation to your Textbook

Interactions between Materials

Then, we see that the glass would have to be positive to attract the negative bakelite.  We might then predict that the (+) glass would then repel the (+) wool (that is , that positive charge repels positive charge, or more generally that like charges repel), and this is in fact consistent with our observations.  We predict that the silk is negative, since it was attracted to the (+) glass, and we finish off by predicting the rest of the interactions:
bakelite (-) should repel silk (-) - correct
wool (+) should attract silk (-)  - correct.
We need to verify this model for many more combinations of materials, but we may well have assured ourselves that it is a valid working model.
Now we have two working models: one which requires two types of charges and two rules, and another which has an infinite number of types of charges (silk-type, wool- type, et c.) and an infinite number of rules.  Which should we accept as correct?  We employ Occam's Razor to assume that the simpler of two equally valid explanations is the correct one.

A Twentieth Century Explanation

In the demos we did in the previous section, we always charged the objects by rubbing them together.  This took charge from one object and deposited it on the other; which way the charge goes depends entirely on energy considerations.  We see that the electrons are almost always what moves, since they are one thousand to ten million times easier to remove from the outer edges of an atom than are the protons deep in the nucleus.  Once removed, they are 2000 times easier to move around, due to their smaller mass.  For example, when we rubbed the wool on the bakelite, electrons were transferred from the wool to the bakelite, causing an excess of electrons on the bakelite, making it negatively charged, and a deficit of electrons on the wool, rendering it positively charged.  This occurred because the electrons have a slightly lower energy when on the Bakelite than when on the wool.

The small amount of work we do by rubbing provides the energy necessary for the electrons to make it over the necessary activation 'hump.'

Properties of Charge

Here is an example to consider:  anti-matter does exist outside of Star Trek^R.  For each sub-atomic particle of matter, there is an anti-particle which is identical in every way except that it has the opposite charge.  When anti-pairs contact one another, both are destroyed (annihilated) and the energy is carried away in electro-magnetic radiation, usually gamma rays.  Consider this reaction:  p⁺ + p^- = 2 g^o.
Is charge conserved in this reaction?

We also say that charge is quantized, by which we mean that there is a smallest non-zero amount of charge possible (given the symbol e), and that all charges are integer multiples of that fundamental charge (i.e., 0, ⁺/-e, ⁺/-2e, ⁺/-3e, et c.).  This is shown by the famous Millikan oil drop experiment (Yes, O.K., earlier work by Faraday suggested quantization).  We find that the fundamental charge amount corresponds to the charge on a proton, which is equal although opposite to the charge on the electron; since the electron was only discovered in 1899 and the proton after that, this is a recent discovery.  Before the discovery of this fundamental quantity, charges were measured in larger, more convenient, but completely arbitrary units called coulombs;  1e = 1.6x10^-19 C, or 1C = 6x10⁺¹⁸ e.

As an aside, there is a theory that there are particles called quarks that have charges less than e (⁺/- ¹/₃e or ⁺/- ²/₃e).  It is thought that combinations of quarks make up protons, neutrons, and many other particles (but not electrons).  This model has been very successful in predicting the results of various reactions between particles, however, individual quarks have never been seen.  For the purposes of this class, they do not exist.

Coulomb's Law

How should we approach situations in which there are more than two charges?  We asserted in Section 4 of the first semestre that forces can be super-imposed, that is, added as vectors.  So we find the force exerted on any given charge by each of the other charged, then add.

Example:
Consider two identical positive charges q which are separated by a distance R; a third, negative charge of the same magnitude is located exactly between the other two.  Find the net force
a) on the leftt hand charge.
b) on the centre charge.

Example:
Three charges are arranged as described below:
 

	Charge	x co-ordinate	y co-ordinate
Q1	-2 nC	0	-0.1 m
Q2	+5 nC	0	0
Q3	+6 nC	+0.3 m	0

Find the magnitude and direction of the electro-static force acting on Q₂.

The Electric Field

Newton worked out, based on Kepler's Laws of Planetary Motion, that two masses M₁ and M₂, separated by distance r, will attract each other with (mutual) forces whose (equal) magnitudes are given by
F = GM₁M₂/r².
Here, G is a constant used just to make the units work out and is equal to 6.67x10^-11 Nm²/kg².  This relationship is Newton's Law of Universal Gravitation.
Let's check it out.  We already know that the weight of an object near the earth's surface is W = g m, and that this same force is described by W = GmM_EARTH/R_EARTH².  So,
g m = GmM_EARTHR_EARTH².
g = GM_EARTH/R_EARTH² = 6.67x10^-11*6x10²⁴/[6.4x10⁶]² = 9.77 N/kg, about what we expected.

Let's consider a fixed charge +Q in space.  If we bring a small test charge +q near to +Q, it will be repelled by a force given by:
F = k_eQq/r².

Let's define the electric field of charge +Q at some location as the force per unit charge that +Q would exert on a test charge q_TEST put in that location:
E = lim_{q ->0} F/q_TEST.
we note that since F is a vector, so is E.  The units of electric field are newtons per coulomb (N/C).
So, if we place +q_TEST at various locations around +Q and find the direction of E at each, we see that the field from +Q points radially outward from +Q.  Can we find the magnitude?  The force between the charges is F = k_eQq_TEST/r², so the field is:
E = F/q = [k_eQq_TEST/r²]/q_TEST = k_eQ/r² radially outward.

How do we find the field from more complicated distributions of charge?  If the distribution is made up of a collection of point charges, then we find the field at some point P by adding the individual fields (as vectors, remember) due to the individual charges.
We did several examples in class.

What if the distribution of charge is not composed of discreet points, but rather is continuous.  In principle, the method is the same as in the last paragraph, but the execution of that method usually requires calculus.  We shall see in the next section a method which will allow us to find E for certain highly symmetrical distributions of charge.  However, there are a few special shapes for which calculus is not required.

Example:
Consider a circular ring of radius R that carries a uniformly distributed charge Q.  What is the electric field at the centre of the ring?

Gauss's Law

Now let's look at a different shape to see if this relationship holds.  If we imagine that the sphere is cut into two hemispheres, the quantity EA will be 2pk_eQ for each half.  Let's shrink one half of the sphere to radius R.

Since EA is independent of the radius, each 'half' sphere will still have flux 2pk_eQ.  However, there is now a third surface to worry about, the flat annular surface which connects the edges of the hemispheres (shown in green).  Here, the area A = p[R² - R²]  and the field  E is certainly not zero (and not even constant!), and so the total flux (as initially defined) is greater than 4pk_eQ.  Rather than give up, can we somehow save our notion?
Notice that the electric field on the hemispherical portions is perpendicular to the surfaces, but that the field along the annulus is parallel to the surface.  If we were to change our definition of flux to E_perpA, the contribution from the annulus would be zero and the total would be
f_E = (E_perp)₁A₁ + (E_perp)₂A₂ + (E_perp)₃A₃= [k_eQ/R²]*2pR² + [k_eQ/R²]*2pR² + [0]*p[R² - R²] = 2pk_eQ + 2pk_eQ + 0 = 4pk_eQ once again.  Yippee!

Now, what if the surface surrounding +Q is of any arbitrary shape?  We can always approximate the surface's shape to an arbitrary degree of accuracy with a series of spherical sections with their centers at +Q.

This gives us two types of individual surfaces dA_i to work with: the curved sections for which E is still perpendicular and for which the flux is (E_perp)_idA_i, and the sides of the sections for which E runs along the surfaces and for which the flux contribution is zero (E_perp = 0).  The total flux through such a surface should be the sum of the fluxes through each small surface, or
f_E = S_i (E_perp)_idA_i.
We can then push and pull the spherical sections until they form a sphere again, remembering that the flux through each individual section does not change as the section is moved, since those that move closer to +Q get smaller areas but have E-field strengths which increase by the same factor that the areas decrease.  In the end, we have a sphere with flux 4pk_eQ again.  So, we can safely assert that the flux from +Q through any closed surface surrounding it must be 4pk_eQ, and that the total flux through that surface can be found from the sums of the fluxes through the component parts of the surface.

What if there is more than one charge?  The total field at any point is the vector sum of the fields due to the individual charges Q₁ and Q₂:
E_total = E₁ + E₂.
As a result, we can also say that the perpendicular component of the total field is the sum of the perpendicular components of the fields due to each charge:
(E_total)_perp = E_1perp + E_2perp.
We can write that
f_total = S_i [(E_total)_perp]_idA_i = S [[E_1perp]_i + [E_2perp]_i]dA_i = S [E_1perp]_idA_i + S [E_2perp]_idA_i = 4pk_eQ₁ + 4pk_eQ₂ = 4pk_e(Q₁ + Q₂),
which we can generalize to as many charges as you like.

What if the charge inside is negative?  Hey, good question, shows you're thinking.  Suppose that I put a positive charge +Q and a smaller negative charge (-q) inside a closed surface, close to each other but far from the surface.  It should be pretty clear that the total electric field at any point on the surface will be reduced from the value it would have had when only the positive charge was there, since the outward pointing field from +Q will be partially if not completely canceled by the inward pointing field from (-q).  This reduction in the magnitude of E_total will correspondingly reduce the total flux
f_total = S_i [(E_total)_perp]_idA_i,
and E_total is smaller than it was).  From the last section, we showed that the total flux should be the sum of the flux due to +Q alone and the flux due to (-q) alone.  Since adding the flux from (-q) reduces the total flux, that number must be negative.  So we define the flux due to a negative charge to be negative.  Now the field at the surface doesn't know or care what causes it, and the flux calculation will be the same regardless, so we generalize the result to say that flux due to any E-line going from the outside of the surface to the inside is counted as negative flux.

What if the charge is outside of the surface?  In that case, the total flux through that surface due to that charge is zero.  Once again, any shape closed surface can be approximated to arbitrary accuracy with the spherical sections centered on the charge.  As above, I can then push and pull the sections to get a nice double spherical surface shape.  Since the quantity EA has already been shown to be independent of the distance from the charge, and since we have just defined the flux from E-lines entering a surface to be negative while the flux from those leaving a surface are positive, and since the fluxes through those side portions of the sections are zero since E_perp = 0, we see that the total flux through this surface due to an exterior charge is zero.

At this point, we have arrived at something useful.

Gauss's Law for Electricity:  The net electric flux through a closed surface is proportional to the net charge inside the surface.

The flux is defined as
f_total = S_i [E_perp]_i dA_i
and it is calculated by looking at each little piece of area, dA_i, multiplying by the component of E_i that is perpendicular to the area, assigning a sign (positive if E points from the inside of the surface to the outside, negative if the reverse is so), and adding up all the contributions from each area.  When this is done, the total should equal 4pk_eQ_{net enclosed}.

Here, we introduce a new and sometimes more convenient constant, e_o, the permittivity of free space.  The value of e_o is 1/[4pk_e].  So, finally, we have that
f_total = S_i [E_perp]_i dA_i = [Q_{net enclosed}]/e_o.

Let's now use Gauss's Law to find the electric field due to some very special, highly symmetric distributions of charge.  The symmetry is important since, although Gauss's law is always true, it is not always possible to work it backwards to find E.

Uniform Infinite Straight Line Charge:

Consider a straight, infinitely long wire which carries a uniform linear charge density l = Q/L.
Let's get an idea what the E-field looks like at some point P a distance R from the line charge

Consider some small bit of charge Q₁, which will cause a field E₁ at P.  For every one of these charges, there is another charge Q₂ the same distance away, but on the other side of P.  The fields E₁ and E₂ will have the same magnitudes, since P is the same distance r from each charge.  In addition, the horizontal components of the two fields will be the same (although in opposite directions) because the angles q are the same.  The resultant E-field from Q₁ and Q₂ is directly away from the line charge.  This is true for any pair of charges, and also for the single charge directly under Point P, so the total E-field at Point P must be radially outward perpendicular to the wire.  What's more, the magnitude of E is the same for any point distance R from the wire, since the argument can be repeated at any spot along the wire.

Now, in order to use Gauss's law, we need a gaussian surface over which to find the flux.  Since the surface is imaginary, we can pick one which will minimize the difficulty of the calculations we must do while still delivering the result we want.  In general, we should pick a surface such that either:
1) the electric field runs along the surface (or part of a surface) so that the perpendicular component of E is zero and the contribution to the flux is zero, or
2) the electric field along a surface (or part of a surface) is constant in magnitude and, if possible, perpendicular to the surface, so that S E_perp dA = EA.

Choose the gaussian surface to be a cylinder of variable radius R and length L, co-axial with the wire.  The surface can be considered to comprise three individual surfaces, the circular end caps and the curved section connecting them.

Gauss's Law: for a closed surface, f_E  = S_i [E_perp]_idA_i = 4pk_eQ_enclosed
The total flux is the sum of the fluxes through each part of the surface:
f_E  = S_{left cap} [E_perp]_i dA_i + S_{right cap} [E_perp]_idA_i + S_{curved part} [E_perp]_idA_i

In that case, the first two terms are zero, since the field runs along the surfaces, not perpendicular to them.  On the curved surface, however, E is perpendicular to the gaussian surface everywhere, so E_perp = E and
S_{curved part} [E_perp]_idA_i = S E_i dA_i

We said above that E is constant in magnitude for any given radius from the centre, so this is true everywhere on the curved surface (E_i = E), so
S_{curved part} E_i dA_i = E S_{curved part}dA_i = EA = E 2pRL = 4pk_eQ_enclosed
Solve for E:
E = 2k_eQ_encl/LR = 2k_e(Q_encl/L)/R = 2k_e(l)/R       (radially outward if l is positive, radially inward if l is negative.)

Note that L dropped out of the result.  This is very important.  Why?

Exactly what it sounds like: a flat sheet that extends infinitely in four directions.  We define s as the amount of charge per unit area:
s = Q/A
We need to get some idea what the E-field looks like before we start.  Use an argument similar to the one outlined above, or use a symmetry argument (done in class) to see that the field must be straight away from the sheet (or towards it, if s is negative).  Also, the magnitude must be the same at any given distance from the sheet, on either side.  Pick a gaussian surface in the shape of a rectangular box (with six sides, top, bottom, left, right, front, back), positioned symmetrically on the sheet, as shown in the figure.

Start with Gauss's Law:  f_E = S_i [E_perp]_idA_i = 4pk_eQ_enclosed  for a closed surface.  Then,
f_E = S_topE_perpdA_top + S_botE_perpdA_bot + S_frontE_perpdA_front + S_backE_perpdA_back + S_leftE_perpdA_left + S_rightE_perpdA_right.

Since E is perpendicular to the sheet of charge, it is along the top, bottom, front, and back surfaces, so that there,
E_{perpendicular} = 0. 
Then, we're left with
f_E = S_left E_perp dA_left + S_right E_perp dA_right

For the same reason, on the left and right surfaces, E = E_{perpendicular}, so that
f_E = S_left E dA + S_right E dA.
Also, E is a constant along each of the right and left hand surfaces and equal in magnitude along both, and the right and left hand sides are the same area as the area of the sheet enclosed by the gaussian surface, so
f_E = E S_leftdA + E S_rightdA = 2 EA
By Gauss's law, f_E = 4pk_eQ_enclosed = 4pk_esA.
So, 2EA = 4pk_esA.
The area cancels out, as we know it should, so that we obtain
E = 2pk_es = s/2e_o.

So, the electric field of an infinite, flat, uniformly charged sheet is constant (althoughthe direction reverses on each side).
Here, we introduce a new and sometimes more convenient constant, e_o, the permittivity of free space.  The value of e_o is 1/[4pk_e] = 9x10^-12 in SI units.

Maxwell developed the notion of the electric field line.  Let's see if we can make some connections to what we've done previously to see if we can suss out any information from their appearance.  Consider the point charge with a radially outward field of magnitude E = k_eQ/r² and the infinite flat sheet with a field of parallel lines with constant magnitude:

We note that in Region A, the lines are close together and the field is strong, while in Region B the lines are far apart and the field is weak.  This correlation seems fine and is consistent with the other example: lines which stay evenly separated (Regions C & D) represent constant field strength.

Following through on this, we can also say that the electric flux through a surface is proportional to the number of field lines passing through it.  For example, we argued that the flux through a spherical section due to a point charge remained constant regardless of r, since the field strength and the area both have r² terms which then cancel in their product, while we can easily see that the number of lines through each surface is also the same.

Since arbitrary shapes of charge can be approximated by a collection of point charges, this should be true for any shape of charge.

Metals

Consider a block of metal; the distributions of electrons and protons are fairly uniform, even to just above the atomic scale (Figure A).

Let's place this block of metal in an external electric field, E_EXT (Figure B, E_EXT is shown in red).

The electrons will experience an electric force (qE) to the left (Figure C, the force is shown in blue), and since they are free to move, they will start to do so toward the rigthhand surface of the metal (they can't go any farther than that).  The protons will experience a force to the left, but each is about 2000 times heavier than the electrons, there are 10 to 100 times as many of them as there are free electrons (plus the neutrons too!), each is locked up inside of an atom, and the atoms are linked to each other in a strong lattice structure, so that we might think of them as composing a single large object, which in turn has macroscopic forces acting from, for example, the table on which the block sits; it's safe to say that the net force on each proton is about zero, so they don't move.
As the electrons move rightward, the separation of charge gives rise to an internal electric field, E_INT (Figure D, induced field shown in green), which points from the excess positive charge on the left toward the excess negative charge on the right.  The net field inside the metal is the vector sum of E_EXT and E_INT, which has a magnitude of (roughly)
E_NET = E_EXT - E_INT.
Since there is still a net field in the metal, even more electrons will travel to the left surface, thus increasing the internal fireld and decreasing the net field.  When does this stop?

So, we see that the electric field inside a conductor in equilibrium must be zero (Figure E); if it were not, then charges would arrange themselves until it is.  Do not be surprised when I qualify this statement in a week or so.

The same argument can be applied to the electric field at the surface of a metal.  Consider an external field which has a component perpendicular to the metal's surface and another along the metal's surface; since charge can move along the syrface, they will once again re-arrange themselves and produce their own field parallel to the surface until the total field has no net component in that direction.  So, we can also say that the electric field at the surface of a conductor in equilibrium must be perpendicular to that surface.

Example 1:
Consider a metal sphere of radius R which has a charge 3Q_o.  Where does that charge reside?

Example 2:
Consider a hollow conducting sphere (inner radius R₁ and outer radius R₂) with a total charge -3Q_o.  Now, place a point charge +2Q_o at the centre of the hollow sphere.  What charge will be on the inner and outer surfaces of the sphere?

What does the electric field look like?  Again consider a concentric gaussian surface of variable radius r.
For 0 < r < R₁,  Q_ENCLOSED = +2Q_o, so E = 2kQ_o/r² outward.
For R₁< r < R₂,  E = 0 (inside metal).
For r > R₂,  Q_ENCLOSED = +2Q_o + (-2Q_o) + (-Q_o) = -Q_o, so E = kQ_o/r² inward.

Example 3:

Consider an infinitely long straight cylinder (radius R₁) of charge with uniform density, that is, the charge is distributed evenly throughout the cylinder.  Let the linear charge density be l₁.  Let a thin cylindrical shell of radius R₂ and with linear charge density l₂ be co-axial with it.  Find the electric field everywhere.
Draw a gaussian surface as a cylinder (variable radius r and length L) co-axial with the real cylinders, and make use of the result of the derivation above for the infinitely long straight wire: E = 2k_el_ENC/r.
For r > R₂, both l₁ and l₂ are enclosed, so E = 2k(l₁ + l₂)/r.
For R₁ < r < R₂, only  l₁ is enclosed, so E = 2kl₁/r.
For 0 < r < R₁, only part of the charge on the central cylinder will be enclosed; we need to figure out how much.  If the charge is distributed evenly within the cylinder, then the fraction of charge enclosed by the gaussian surface is the same as the fraction of the volume enclosed:
l_ENC/l₁ = pr²L/pR₁²L,
so that
l_ENC = l₁r²/R₁².
Substituting gives
E = 2kl_ENC/r = 2k[l₁r²/R₁²]/r = 2kl₁r/R₁².

Example 4:
A point charge of +3Q_o is located at the centre of a hollow sphere (inner radius R₁, outer radius R₂) that has a uniform charge density throughout and a total charge of +Q_o.  Find the electric field everywhere.

Dipoles

There is some potential energy associated with the dipole, since the kinetic energy of this motion had to come from somewhere (we can think of the source either as PE or as the work done by the electric force, but never as both).  That PE is
PE_dipole = -pEcosq _p,E;
let's see if we can derive this result.
Start with the dipole arranged such that (q _p,E )_o is 90^o.

Now, let the dipole rotate to an arbitrary angle q _p,E.

To calculate the work done by the electric force on, say, the positive charge, we need only consider the displacement in the direction of the force, which will be [L/2]cosq _p,E.  The work done is then
[F₊][L/2]cosq_p,E = [qE L/2]cosq_p,E = ¹/₂ [qL E] cosq_p,E = ¹/₂ pE cosq_p,E.
A similar amount of work is done on the negative charge, so that the total work is
W_E-field = pE cosq_p,E.
But, we defined the change in PE to be
DPE_dipole = - W_E-Field = - pE cosq_p,E.
Now, if we set the PE to zero when q_p,E = 90^o (and why not?), then we have our result, that
PE_dipole = - pE cosq_p,E.

Mastery Question

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