Student questions from Cook Chapter 140:

1. Could you explain the calculation of a dipole in example 140.2? I am confused by the use of "p."

Obtaining equation 140.9, the electric field of a dipole, is not straight forward. You can try working through P140.5 to get there. You have to use the binomial expansion to approximate the expression for the field for locations that are far from the origin compared to a.

2. On page 14005 Cook is trying to explain calculating the field of point charges. I dont understand those equations at all or the list of variables shown above them.

I went through this example from a slightly different perspective in class. The list of variables in the middle of page 14005 are the q_i and r_i 's that go into equation 140.6 for the electric field due to a collection of point charges.

3. Please walk through how to get the answer to E140.1. I don't understand how to deal with i and j

Use equation 140.5 with r'= 0 (there is only one source charge and it is located at the origin). So the x-component of the electric field at (x,y)=(3m,4m) is qx/(4*pi*eps_o*r^3) where r^3 = (x^2+y^2)^3/2. The y-component is qy/(4*pi*eps_o*r^3). Whenever you encounter a vector equation, you may break it up into three equations, one for each (mutually perpendicular) component. When we write down a vector equation we mean that the x-components of the left side equal the x-components on the right side. The y-components on the left side equal the y-components on the right side, etc.

4. could you possibly do an example for an electric dipole moment in class?

I do not know if I will have time to do such an example. You had such an example on the last problem set though, P130.2. The electric dipole moment p for that problem is qa where the vector a has magnitude equal to the distance between +q and -q and is in the direction from -q toward +q.

5. I don't understand the scattering path was given in figure 135.5, do they get these experimentally or do they calculate them?

Those trajectories are calculated. They are the hyperbolic trajectories that are the same as the solutions to the gravitational motion problem. Note the similarity (identity!) of figures 135.5 and 80.11. In the gravitational case, These orbits are the unbound orbits typical of comets. In the electric case, they are the scattering trajectories for a light charged particle passing nearby a heavy charged particle.

6. Can example 140.1 be done using the equation kq(/r^2) * dr?

Your equation is a little ambiguous. You need to be careful to keep track of the vector nature of the contributions to the total electric field due to the individual "source" charges. It is not clear from your equation how you are dealing with those vectors.

7. I'm sure you'll answer this in class, but what is the answer to E140.8?

I did not get to the point of discussing electric potential in class. The answer to E140.8 is NO. In order to determine the electric potential at some point in space, you need to know the electric field at every point along some path from your reference location to the point of interest. You need to perform the path integral expressed in equation 140.19.

8. Most of chapter 140 was making sense until I reached Section 140.3.3 on pages 14007-14010. I'm pretty shaky on any kind of integration, and this one pretty much lost me - especially the "new integration variable." To begin with, I really can't understand the "what" and "why" of Example 140.3. Once I have a grasp of that, I'll probably better know what other questions to ask.

Example 140.3 is essentially the same example I did in class. The "what" is to calculate the electric field due to an infinitely long charged rod. The "why" is becuase, although one can never have a truly infinite straight wire, there are real situations which approximate this case... and because it is one of the relatively small number of charge distributions for which we can write down an "exact" solution for the electric field everywhere.

If you can understand equation 140.6... the electric field due to a collection of point charges located at positions r1,r2,r3, etc, then equation 140.11 is really the same equation. The sum becomes an integral when the number of point charges to be added up becomes large and they are close enough together that one can approximate the collection of point charges by a smooth distribution of charge density throughout some region of space. I suggest concentrating on the examples (140.3, 140.4, and 140.5) to get a feel for how equation 140.11 is put into application.

9. The more I read, the more I realized that I don't know anything about path integrals. Can you explain the mathematics of them, if only briefly?

Go back to chapter 110 and try to make sense of Fig. 110.3. That figure illustrates the path integral involved in the calculation of the work done by a force. You are probably having trouble because in your calculus experience, you have only had to deal with integration of functions of a single variable. Now, we are considering functions of three variables (x,y, and z). There are several different ways to integrate a function of three variables. One can integrate it along a single dimension that coincides with one of the coordinate axes. That is a special case of the path integral where the path lies along the coordinate axis. More generally the one dimensional integral can be along a path that winds its way through three-dimensional space. One can also integrate a function of three variables over a two dimensional surface. Finally one could integrate the function over a three-dimensional volume.

10. I've also been thinking about a physics problem that seems easy, but seems continually as one puts more thought into it. If there is a spacecraft between the earth and the moon, it has gravitational influences from the earth and the moon. Yet, the spaceship has mass and so exerts gravity of its own on both bodies. Is there a general equation for the force exerted by all three objects? I've been thinking along the lines of the F=(GMm/r^2) formula we learned (but this formula only relates two masses). This question assumably has an answer as engineers would need to know this general formula to decide how much thrust a spacecraft would need to arrive at a particular place at a particular velocity, etc.

Let's be careful about the object we are interested in and the forces acting on that object alone. We do not consider an object to exert a force on itself (although in truth different parts of an object can exert forces on other parts of the same object. These forces however are internal forces and cancel out due to Newton's 3rd law when considering the net forces on the object). You are contemplating a spacecraft with gravitational forces acting on it due to the earth and the moon. Well you use the law of universal gravitation to determine the VECTOR force on the spacecraft due to each body individually and then you add the vectors to get the net force on the spacecraft. We are doing the same thing when we consider the electric field at a point in space due to multiple electric charges at various locations in space.

11. In Cook 15015, it is stated, "In many cases where constant B-fields are needed, Helmholtz coils are a natural selection as a source." What is such a case?

You are a little ahead of us if you have already read chapter 150. The answer to your question is that Helmholtz coils are easy to construct and provide a region of very uniform magnetic field. There are interesting and important effects that occur when atoms are placed in magnetic fields. To observe those effects cleanly, one would like the entire sample of atoms being studied to have the same magnetic field applied to them.

12. I like to think of things concretely -- what physically happens under certain conditions (the formulas don't mean anything to me by themselves). Are there concrete ways to think of fields? Specifically, what exactly are electric and magnetic fluxes?

Formulae never mean anything by themselves. You have to connect them to operational definitions of concepts and to explicit examples where those equations and concepts are used to solve problems. Ultimately you cannot avoid coming to grips with the equations however and "seeing" the concepts that those equations summarize.

Sometimes, with very abstract concepts it helps to have an analogy to ground your initial understanding of the concept. Let me try one on you that might help you understand electric and magnetic fluxes. The electric field (and the magnetic field) is a vector field. Another, more concrete vector field is the flow velocity of the water in a river. At each point in the river there is a vector that indicates the velocity (magnitude and direction) of the water at that point in space. Now take a small hoop and place it in the river. Determine the number of kilograms of water that flow through that hoop per second. That number... (X kg/s) is related to the "velocity flux" through the hoop. Clearly it depends on the orientation of the hoop in the stream You can orient the hoop so that the water does not flow through it... when the normal to the plane of the hoop is perpendicular to the flow. The velocity flux through the hoop is calculated as the component of the velocity perpendicular to the plane of the hoop (or parallel to a vector normal to the plane of the loop... dS), times the area of the hoop. To get the number of kilograms of water per second passign through the hoop you would need to multiply your answer by the density of water. If you wanted to determine the velocity flux (or the mass per unit time) passing through a larger surface over which the direction and magnitude of the velocity varied... you would have to break the surface up into small pieces (little hoops), calculate the flux through each of the small pieces, then add them up. I just described a surface integral of the flux. If you consider the flux (of water) through a closed surface (a sphere perhaps) you can determine the number of kilograms of water leaving a closed region per second. If you get a positive number, then either the mass of water remaining inside the surface is decreasing OR there is a source inside the closed region.. a spring or a faucet.

You can use this analogy to consider the electric flux through a surface (or the magnetic flux through a surface). If the vector field associated with the electric field were replaced by an identical velocity field for a flowing fluid then the flux would be related to the number of kg/s of fluid flowing through the surface. The net electric flux through a closed surface can only be nonzero if there is a "source" of electric flux inside the closed surface. Electric charges are sources for electric flux. This is the essence of Gauss' Law. It is important to keep in mind that nothing is moving in the case of the electric field. The fluid analogy is an analogy, not an equivalence.

13. How does B shape figure 135.4.

I am not sure what you mean by "shape." Without he magnetic field (which is directed out of the page in that figure) the (positively) charged particle would accelerate in the y-direction. If the magnetic field were present but without the electric field, the particle would gyrate with (clockwise) circular motion at the cyclotron frequency (Eq. 135.27). With both fields present (both uniform and perpendicular to each other) the particle gyrates, but the center of gyration drifts in a direction perpendicular to both E and B.

14. For E140.1, I got the same numerical answers, but I got a negative number. The charge is negative, so in order for the answer to be positive, vector R would have to be negative also--but isn't vector r = 3i^+4j^ (and therefore vector R is positive)?

You are correct. The electric field point toward a negative charge. The numbers are correct, but should be negative.

15. I don't understand why field lines would be tangents--what is it about the equation 140.6 that tells us that they are essentially derivatives?

We choose as the most logical and informative way to construct field line maps the following (arbitrary) rules:
1) the actual field (electric or magnetic) is tangent to the field line at any point, and
2) the magnitude of the field is proportional to the number of field lines per unit area at right angles to the field.
Equation 140.6 does not tell you that the electric field is a derivative...rather the logic goes this way:
1) The (static) electric force is a conservative force, therefore we can define a potential energy associated with it.
2) Since the electric force on an object is proportional to its charge, we can define the electric field as the electric force per unit charge... and the electric potential as the electric potential energy per unit charge.
3) It follows that the direction of the electric force is always perpendicular to surfaces of constant electric potential energy AND can be determined by the appropriate combination of partial derivatives in the x, y, and z-directions (the gradient) of the electric potential energy.
4) The electric field is also therefore perpendicular to surfaces of constant electric potential and is determined from the gradient of the electric potential.

16. What did Cook do at the top of page 14009? I think he held zeta constantto evaluate the integral, but then if the z component is 0, why isn't the whole integral 0? Are the homework problems going to be as difficult as this one?

A vector equation is really three equations, one for each component. Cook argues that the z-component of equation 140.15 is zero. Read the footnote (#4) at the bottom of page 14009 for an explanation. The argument does NOT apply to the x and y components of equation 140.15. Those two component equations involve identical integrals and are solved in equation 140.16.

17. On page 14009, Cook says that -xi^+zk^ is perpendicular to the source and extends from source to the observation point, but isn't the plane the source? If so, the vector wouldn't be perpendicular, especially in the case drawn.

Cook uses the solution from example 140.3 to help solve example 140.4. The source referred to in the paranthetical statement 2/3 the way down page 14009 is the element of the total source that correspond to the line charge from example 140.3. You have to add up contributions to the electric field from an infinite number of line sources to get the total electric field due to a plane of charge. The vector -xi + zk is perpendicular to the strip (line) at x extending from minus infinity to plus inifinity in the y-direction. The contribution of a single strip to the total electric field is the same as the electric field due to an infinitely long charged line. The total electric field is found by adding up (integrating over) the contributions of the strip elements... making sure that the vector nature of the sum is taken into account.

18. To calculate the integral on the top of page 14010, Cook says to substitute s = z*tanx, but what are we substituting it for? I don't see an s anywhere in the equation.

If you have not completed Calc II yet, you may not have encountered the trig substitution technique. Or, you may be confused because the substitution suggested is not quite the right one. Make the substitution x = z tan q in the equation for the z component of E at the bottom of page 14009. So the denominator is z^2*(1 + tan^2 q). The numerator (dx) must be found by differentiating the z tan q with respect to q. You get dx = z*(1 + tan^2 q) dq. So after substitution, the integral becomes integral of dq from 0 to (pi/2). Remember to transform the limits of integration when you change variables. That gets you to the equation at the top of p. 14010.

19. Could we go over E140.5 and 140.6 in class?

Yes. I will go over this. The electric field lines for an infinite charged sheet are straight and perpendicular to the sheet. The line begin on the sheet (if it is positively charged) and go straight off to infinity. For two inifinite charged sheets... one positively charged, the other charged negatively but of equal magnitude... the electric field lines begin on the positively charged sheet and end on the negatively charged sheet.

20. Where did the k^ go in the E_z equations of Example 140.5?

Only the z-component of the electric field is shown. The other two components are zero, but since the equation specified only one component, the unit vector need not be shown.

21. E_z, both, equation: I understand where the top and bottom came from, but not the middle. Though it is intuitively understandable, I'd like to know where it came from mathematically.

The total electric field at a point in space is the superposition of the electric fields due to all charged present. You have two infinite charge sheets, one positive and the other negative, and the two sheets are separated by a distance d. The total electric field above both sheets is zero because the fields due to the individual sheets is the same magnitude but in opposite directions. The same is true below both sheets. However, between the two sheets the fields add (they are in the same direction as well as being of equal magnitude).

22. I was looking through the problems and noticed that P135.7 stated that an electron and a proton both have the same kinetic energy and move in circular paths in the same magentic induction field. I don't understand how that could be. First of all wouldn't the proton be attached to the neutron in the nucleus and would be the positive that would create the electron the negative to stay around the nucleus? So how can a proton and an electron be moving in a magnetic induction field?

The proton and the electron are assumed to be "free" in this problem. Electrons can be liberated from atoms and allowed to move freely in space. If the electron comes from a hydrogen atom, the liberated electron leaves behind a free proton (the nucleus of the hydrogen atom). Consider the motion of each of these "free" particles with equal kinetic energies in the presence of a magnetic field. Remember that the proton and the electron have charge of equal magntidue but opposite sign, while they have very different masses. 

23. I don't understand equation 140.30.

The electric potential difference between two locations is a (path) integral over the (vector) electric field. Equation 140.30 derived the reverse relationship... that the electric field is a derivative (actually the gradient) of the electric potential. Cook gets at that relationship by considering the work done (by the electric force) on a charged particle in moving a small distance in the x-direction. For a very small distance the work done is the component of the force in the direction of motion (qE_x) times the distance (delta_x). But the work done by a conservative force also equals minus the change in potential energy (delta_Phi_e = q*delta_phi, where Phi_e is the electric potential energy and phi is the electric potential). Solving for the x-component of the electric field, one gets that E_x = - d(phi)/dx. If the electric potential depends on y and z as well (as it generally does), then the x-component of E is found using the partial derivative of phi with respect to x. One can perform the same steps with the charged particle moving in the y or z-directions to get the y and z-components of the electric field. The final result is that the electric field vector is minus the gradient of the electric potential (Eq. 140.32).

24. Flux is the sum of what?

See answer to question #12.

25. On E140.10, as with E140.1, I got the right numerical answer but the wrong sign. How does Cook get a positive answer while using a negative charge?

26. On page 14017, how are we supposed to calculate the field in step 2? Also, how do we know that the equipotential surfaces are spherical (or circular in 2D)?

27.Where did equation 140.36 and 140.37 come from? 140.37 makes sense if R^ is parallel to dS, but how can a length be parallel to an area?