Having finished our discussion of DC circuits (Chapters 25 and 26), we now return to the study of the details regarding the force between two charges, the force experienced by an electric charge when placed in an electric field, and the calculation of the electric field that surrounds an electric charge (Chapter 21 sections 3, 4, and 5).

We then cover an easier and more elegant way to calculate the electric field using Gauss's Law (Chapter 22) and then the important topic of Electric Potential (Chapter 23). The electric potential (given in units of Volts) is smoothes reefed to as the voltage, and we will see that we can also easily calculate the electric field from the electric potential.

KEY TERMS: point charge, Coulomb's Law, coulomb, principle of superposition of forces, electric field, test charge, vector field, source point, field point.

By careful laboratory experiments, Coulomb in 1784 deduced that the magnitude of the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them:

 |F| = k |q1 q2| / r^2                or                                

Here the carefully measured proportionality constant is (as far as we are concerned):

k = 1 / (4p e_o) = 9 (10)^{^9} N m^2/ C^2

( e_o is called the permittivity of free space, and we will discuss this further in our study of capacitors and dielectrics in Chapter 24.)

The direction of the force (vector) is always directed along the line joining the two charges, and when the charges have the same sign the force is repulsive and when the signs are opposite the force is attractive. 

What is the total force exerted on the charge q₃ by the other two point charges?
a) The three charges. (b)Free body diagram for q3.

Vetor addition of electric forces on a line.

q3 = 1.0 nC, q2 = 3.0 nC, q3 = 5.0 nC   (n = nano = 10^-9)

F1 on 3 = 112 uN, F 2 on 3 = 84 uN, and the total force on
q3 is (84 - 112) uN toward the left
(u = micro = 10^-6).

Since there is symmetry about the x-axis the vertical (y-axis) components of the two forces cancel and we only need to calculate the horizontal forces.   
F 1 on q = k q1 Q / r^² = 0.29 N
and the horizontal
(x-component) =
0.29 N cos a =
0.29 N (0.40/0.50) = 0.23 N.
The total force on Q is 2(0.23 N) = 0.46 N to the left (+x-axis).

F 1 on q is the force on Q
due to the upper charge q₁.

We know that there exists a Coulomb force of repulsion between the two charged bodies at A and B, (having charges Q and qo respectively) whose magnitude is: 
F = k |Q qo| / r^² =
qo [k Q / r^²] where we have factored out the small charge qo.
 We can write the force in terms of an (electric) field having units of force per unit charge, or Newtons per Coulomb, so that:  F = qo E.  Therefore the electric field
E = [ k Q / r² ]
which has units of N/C. 

(This is very similar to Newton's gravitational force F = G mM / r^{^2} = m [G M / r^{^2}] = m g where the gravitational field, g, has units of force per mass, or Newtons per kilogram, which equals mass per seconds squared, an acceleration of 9.8 m/s^{^2} near the surface of the Earth.)

Electric field at three points, a, b, c, set up by charges q₁ and q₁, which form an electric dipole.
The electric dipole's charges are q1 = +12 nC and q2 = -12 nC

Calculate E₁, E₂, and total field E_T at points a, b, c.

At point "a"
E₁=3.0 (10)^4 N/C,
E₂=6.8 (10)^4 N/C,
E_T=9.8 (10)^4 N/C, all in +x direction. 

At point "b"
E₁=6.8 (10)^4 N/C,
E₂=.55 (10)^4 N/C,
E_T=-6.2 (10)^4 N/C
all in +x direction. 

At point "c"
E₁=6.4 (10)^4 N/C,
E₂=6.4 (10)^4 N/C,
E_T=4.9 (10)^4 N/C,
E_T is in the +x direction. 

Calculating the electric field on the axis of a ring of charge. 
In this figure, the charge is assumed to be positive.

The ring has a total charge of Q distributed uniformly around it.  Find the electric field at a point P that lies on the axis of the ring at a distance x from its center. 

The magnitude of dE = k dQ/ r^{^2} and the x-component is
dEx = dE cos a = {k dQ / x² + a² ] {x / [x² + a²]^{^1/2} }. 
From symmetry we see that all the y- and z- components of E cancel and only the x-components do not cancel.  Now get the total x-component of the field by adding up the contribution of all the dEx's (i.e., by integrating): 
Ex = _ò  k x dQ / [x² + a²]^{^3/2}
where x is constant as we go from point to point in the integration, so Ex = k x Q / [x² + a²]^1/2

Fig. 21.21 Electric field on the axis of a ring of charge.

Problem 21.86:  Positive charge Q is distributed uniformly along the positive x-axis between x = 0 and x = a.  A positive point charge +q lies on the positive x-axis, a distance x = a + b, a distance b to the right of the end of Q.  (a) Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis where x > a.  (b) Then calculate the force (magnitude and direction) that the charge distribution Q exerts on q. 

Note the tabulated integral: ò dz/(c-z)^2 = 1/(c-z)

Problem 21.87:  Positive charge Q is distributed uniformly along the positive y-axis between y = 0 and y = a.  A negative point charge -q lies on the positive x-axis, a distance x from the origin.  (a) Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis.  (b) Calculate the x- and y-components of the force that the charge distribution Q exerts on q. 

Tabulated integral: ò dz /(z² + a²)^3/2 = z / a² (z² + a²)^½

Now that we know how to calculate the electric field from a distribution of electric charges, we can look at another way to do the calculation which is much more elegant and much more simple - the integrals will be trivial to perform. The method is called GAUSS'S LAW and is discussed in Chapter 22.

GAUSS'S LAW (Chapter 22)

Using Gauss's Law to calculate the electric field caused by a line of charge or other distribution of charges will be much easier than dealing with the messy integration we used in the previous chapter. In fact, the integrals will be so trivial we can literally do them in our heads!  Here's how we use Gauss's Law:  Surround a given distribution of electric charges with an imaginary surface that totally surrounds the charge.  Gauss's Law gives the relationship between the electric field at all points on the imaginary surface and the electric charge enclosed by the surface.  In practice, this is very easy to use to determine the electric field caused by the charges inside the surface. 

Check the Summary and these "Key Terms" at the end of the chapter: 
closed surface, electric flux, surface integral, Gauss's Law,
Gaussian surface.

Section 22.1 Charge and Electric Flux is worth reading slowly and carefully.  It gives the answer to the question "If the electric field pattern is known in a given region, what can we determine about the charge distribution in that region?"  Knowing this we will use the law to calculate (easily!) the electric field caused by a given distribution of electric charges. 

For a clear picture of Gauss's Law go to an informative website with interactive pictures that will quicky show you pictures that will help give you a better idea of what Coulomb's Law is all about:

http://230nsc1.phy-astr.gsu.edu/hbase/hframe.html

then click on "Electricity and Magnetism" button,
then "Gauss's Law" button and then Gauss's Law Applications." Check out the several applications using the buttons at the bottom of the page for calculations you may see on a test later.

You can click anyplace on the interactive page to get an explanations of the many new ideas in this chapter. You will find this very helpful to have a picture of some of these abstract and complicated concepts.

The Figure (22.2) shows that the pattern of the field on the surfaces of the box enclosing the charges is different depending on what charges are enclosed.  For example, when the charges are positive the field is outward, when the charges are negative the field is inward, and it is clear that if there were no charges the field on the surface of the box would be zero.  If the amount of charge inside the box increases, the field pattern on the surface of the box changes accordingly. 

The electric field vectors surrounding electric charges are considered to be a vector field, similar to the vector field of the wind patterns we sometimes see in a weather forecast.  We can visualize the flow of the wind by looking at the wind velocity (field) vectors.  Although the electric field does not flow in any physical sense, we can think of the electric field vectors to be analogous to the velocity (flow) vectors on a weather map, and we call the "flow" of the electric field vectors the electric flux (from the Latin word flux which means flow).  This concept of electric flux gives us a convenient way to visualize the electric field vectors. 

Looking at Figure 22.2 we can now call the outward pointing electric field vectors an outward electric flux through the box's surface, and the inward pointing vectors an inward electric flux through the box's surface.  If we had one positive and one negative charge in a box the net flux through the surface of the box would be zero because the outward electric field vectors would cancel the inward pointing vectors.  The zero flux correlates with the zero net electric charge in the box. 

If we had a charge outside the box it would contribute a zero net flux through the box surfaces because every electric field line from the charge that entered the box would have to exit through the other side of the box.  This is an important result -- we need only consider charges inside our imaginary box when calculating the electric field vectors on the surface of the box. 

Since the magnitude of an electric field vector is directly proportional to the magnitude of the electric charge, the electric flux through the box surfaces will be directly proportional to the magnitude of the strength of the charge inside the box. 

To summarize:

• The sign of the enclosed charge determines the direction (in or out) of the electric flux through the surface of the box. 
  
  
• Charges outside the box do not contribute to the net electric flux through the box.
  
  
• The net electric flux through the box surface is proportional to the magnitude of the electric charge inside the box. 

The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to the velocity vector v and (b) is VA cos f when the rectangle is tilted at an angle f.

We will next replace the the fluid velocity flow vector v with the electric field vector E to get the concept of electric flux F_E.

(a) The electric flux through the surface equals EA.

(b) When the area vector makes and angle f with the vector E, the area projected onto a plane oriented perpendicular to the flow is A cos f.  The flux is zero when f = 90 deg. because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle.

Let's define an area vector A to have a magnitude equal to the area of a surface and a direction perpendicular to the area of the surface.  Then we can write the term VA cos f more concisely as a dot product of two  vectors:  v . A  = VA cos f.  If we now replace the fluid velocity flow vector v with the electric field vector E we get an expression for the electric flux:  F_E = E . dA = VA cosf.

If the electric field vector E varies from place to place over the area we simply divide the area into small pieces dA which is a vector with magnitude dA and direction perpendicular to the surface ( and direction outward if the surface is curved) to get:

F_E =ò E.dA = ò E dA cos f 

a general definition of the electric flux.  This is a surface integral in which the value of E at each point on the surface is multiplied by the surface area element dA at that point and integrated (or "added"). 

Consider Gauss's Law and the electric field caused by a point charge q.  (We know from last chapter that the magnitude of the field vector is
E = (4pe_o) q /R^{^2}). 
Place the positive charge q at the center of an imaginary spherical surface whose radius is R. 
F_E = ò E.dA = ò E dA cos f =
ò E dA = E ò dA =
E (4pR^2) = (4pe_o) q /R^{^2})(4pR^2) = q / e_o. So we have the
electric flux  F_E = q / e_o. 
And if we have more than one charge inside the sphere we simply add them up because each charge will contribute to the vector field E.  Therefore the total charge enclosed by the imaginary sphere is Qencl = q₁+ q₂ + q₃ + ...
Now we can write Gauss's Law:

F_E = = ò E.dA =   = ò E dA cos f = Qencl/ e_o

The projection of an element of area dA of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by two, so the area element on the larger sphere is 4 dA The same number of line of flux pass through each area element.

If the conductor is isolated (not connected to a battery, etc.) all motion of excess charges will stop after the charging process has taken place.  By excess charges we mean those charges other than the ions and free electrons that the neutral conductor is made of.  So if the situation is electrostatic there can be no electric field inside the conductor.  If there were a field it would exert a force on the charges and they would move. 

If we now draw our imaginary Gaussian surface completely inside the conductor as shown in Figure 22.17 there can be no electric field on the imaginary surface because it is inside the conductor.  By Gauss's Law we must have Qencl = 0. We conclude that all the excess charges must reside on the outer surface of the solid conductor. 

This is a very important result that we will use repeatedly in our calculations of electric fields by Gauss's Law:  There are no excess charges in the interior of a solid conductor, i.e., Qencl  = 0.

Use the following recipe for Gauss's Law types of problems:

1. Carefully draw a figure showing the location of all charges, the direction of the resulting electric field vectors E, and the dimensions given.

2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface (or a portion of the surface) and the surface contains the point at which you want to calculate the field.  This is very important because you will not be able to do the integral if the magnitude of E cannot be taken outside the integral in Gauss's Law.

3. Write Gauss's Law and perform the dot product E . dA on the vectors to get rid of the vectors (vector calculus is difficult). 

4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the E out of the integral and then easily perform the integral.

5. Determine the value of Qencl from your figure and insert it into Gauss's equation.

6. Solve the equation for the magnitude of E.

If a conducting sphere of radius R has a positive charge +q, we can calculate the field inside and outside the sphere by using Gauss's Law. 

1. Inside the conducting sphere the electric field is zero. 
(We have already shown this.)

2. To calculate the field outside the sphere draw the figure showing all the charges, indicate the direction of E at the point (r) you want to calculate E, and let r > R. 

3. Do the dot product in Gauss's Law: 
ò E.dA = ò EdA cos f = Qencl /e_o. Evaluate cos f = cos 0 deg =  1

4. Factor E out of the integral and integrate
E ò dA = E (4p R^²) = Qenc /e_o).

5. Insert Qencl = +q (see Gaussian surface) and solve for E to get:   
E = +q / e_o(4p R^²)  

Field of a Line Charge - Electric charge is distributed uniformly along an infinitely long, thin wire.  The charge per unit length is +l (in Coulombs per meter).  Find the electric field at a distance r perpendicular to the wire from the center of the line (see figure). 

1. Draw the figure showing all the charges, indicate the direction of E at the point (r) you want to calculate E, and let r > R.  The symmetry of the charge distribution indicates that the resultant electric field vector will be in a plane perpendicular to the axis and radially outward. 

2. The geometry of the charge distribution suggests that we draw a cylindrical closed surface containing the point at r, the distance from the charges.  The direction of E will be perpendicular to the curved portion of the cylinder and outward. 

3. Write the integral over the closed surface in three parts - an integral over each flat end and an integral over the curved surface.  Write Gauss's Law with the three integrals and do the dot product in each: 
ò  E1.dA1 + ò  E2.dA2 + ò  E3.dA3 =
ò  E1 dA1 cos f1 + ò  E2 dA2 cos f2 + ò  E3 dA3 cos f3 =
0 + 0 + ò E3 dA3

The zeros result because the angle between the E vectors and the area dA vectors of the end caps of the cylindrical surface is 90 deg. and cos 90 ^deg. = 0.  The angle between the E vectors and the area vector of the curved surface is 0 ^deg. , and
cos 0 ^deg. = 1.    (Let's discuss this in class!)

4. So Gauss's Law becomes ò  E.dA =ò EdA = E ò dA =
E (2prl) = Qencl / e_o, where (2prl) is the area of the curved Gaussian surface.

5. Now evaluate by noting that the "linear charge density" is l = charge / length  or  charge = l (length) or Qencl = l l where l is the length of our cylinder. 

6. Solving for E we get    E = l l  / e_o  (2prl) and the l's cancel, so  
E = l / e_o (2pr).    

Field of an infinite plane sheet of charge - Find the electric field caused by a thin, flat, infinite sheet on which there is a uniform positive charge per unit area s.

The nature of the charge distribution suggests that our Gaussian surface will be a cylinder with its axis perpendicular to the plane of charge.  We need to write the integral in Gauss's Law in terms of three integrals over the three parts of our closed cylindrical surface just as we did in the problem above for the long charged wire.  When we evaluate the dot products in the three integrals we will find that one of them goes to zero because the cos 90 ^deg. = 0.  When we evaluate the term Qencl we use the definition of the "surface charge density" s = charge / unit area = Qencl  / A
where A is the area of the end caps of our cylinder.  We get Qencl  =  sA and the rest is easy.    (Let's discuss this in class!)     The answer is E = s / 2e_o.

There is an electric field between oppositely charged parallel conducting plates, i.e., a capacitor -- the subject of Chapter 24.

Two large plane parallel conducting plates are given charges of equal magnitude and opposite sign; the charge per unit area is +s for one and -s for the other.  Find the electric field in the region between the plates. 

The nature of the charge distribution suggests that our Gaussian surface will be a cylinder with its axis perpendicular to the plane of charge.  We need to write the integral in Gauss's Law in terms of three integrals over the three parts of our closed cylindrical surface just as we did in the problem above for the infinite plane sheet of charge density s.  We will wisely place one of the end caps of our Gaussian cylinder inside the conducting plates.  When we evaluate the surface integral for this surface we can make use of the fact that we know the electric field inside a conductor is zero and the integral will nicely go to zero!  
(Let's discuss this in class!)    

The field outside the sphere is the same as that for a solid conducting sphere with the same amount of total charge but spread over its surface.  The field inside the sphere increases linearly with r, i.e., the field gets larger as we go further away from the center and our Gaussian spherical surface encloses more and more charge.  We need to characterize the charge distribution causing a "volume charge density" r = charge / unit volume.       
We will omit this problem!

(a) Charge on a solid conductor resides entirely on its outer surface.
(b) If there is no charge inside the conductor's cavity, the net charge on the surface of the CAVITY is zero.
(c) If there is a charge +q inside the cavity, the total charge on the cavity surface is -q.

The solution of this problem lies in the fact that the field inside a conductor is zero and if we place our Gaussian surface inside the conductor (the field is zero), the charge enclosed must be zero, i.e., +q and -q = 0 charge enclosed by the Gaussian surface.

(a) When a positive charge moves in the direction of an electric field, the field does positive work and the potential energy decreases.
Work = qo E d

(b) When a positive charge moves in a direction opposite to an electric field, the field does negative work and the potential energy increases.

In both cases (poinitive or negative piont charges), if you move in the direction of E, electric potential V decreases; if you move in the direction opposite E, V increases.