Lesson 2:

Electric Force, E field, Energy and Voltage: Know the Patterns

Read With a Critical Mind

While you have recently seen that there are two new formulas that say F = kqQ/r2for force and E =kQ/r2for electric field, it does not necessarily follow that these are always used to calculate electric field. The Lesson 2 notes look at electric field from a mathematically simpler and more conceptual point of view, and the two formulas just mentioned will be mostly absent (even though they are still true.) So why should you care about electric field? It’s my guess that all you know is that voltages are easy to measure and that you are taking my word for it that electric field has something to do with voltage. That is sufficient background knowledge for now. Follow directions now.

First Exercise: Basic Direction of E field

Imagine a proton placed at point R below. In which direction would it feel a force? _____

That was easy. It’s leftward (repulsion.) But now I don’t want to talk about force. I want to talk about something called Electric Field, AKA E field. Go read what your book in Chapter 17 says for the definition of Electric Field, and then report back here.

+ –

Universe A: Universe B:

R S T P

+ –

The flat plate in universe A is positively charged. The one in universe B is negatively charged. Things in Universe A do not affect Universe B. For ALL of the points (R, S, T, P) near the charged plates:

  1. State the direction of the force felt by a positive test charge placed at each lettered point. (Example: at point R, the force felt is to the left.)
  2. State the direction of the force felt by a negative test charge placed at each lettered point.
  3. Use the definition F = qE to determine the direction of the E field at each lettered point when there is a negative test charge there. This is not supposed to be obvious, so the exercise is done at point R as an example for you now:

Solution: Consider point R. When a – test charge is there, the force it feels is to the right. The test charge is negative. Therefore substitution into F = qE will allow me to find the direction of E:

F = (q)(E)

(right) = ( – ) (direction of E)

Answer: the direction of E at point R is to the left, because the negative of left is right. Just like the negative of southwest is northeast.

3. Now you figure out the direction of the E field at the other three points.

  1. Now use F = qE to find the direction of the E field at points R, S, T, and P supposing that there is a positive test charge located at each point.

Please pay close attention to what you just figured out: Look at the E field directions you answered in questions 3 and 4. Do these answers for the E field depend on whether the test charge at the point is positive or negative?

Everything so far tells us that the E field is only dependent on the source charge, and we say that E field exists in the space around the source whether there is any test charge in the space to feel it or not. This is what you should have learned from these notes so far.

E field lines originate on and point away from ______charges, and they converge upon and points towards ______charges.

Figuring Out the E field of a parallel plate capacitor

In the diagram below, use pencil to draw the E field due to the positive plate. Your E field must cover ALL space as if the negative plate were not there. And make sure you show its direction with arrowheads.

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Now draw the E field due to the negative plate. Cover ALL space as if the positive plate were not there. Now erase the E field lines in region where opposite directions cancel. Look what you are left with. It is the E field inside the plates of a parallel plate capacitor. And outside the capacitor, there is no E field. Unlike gravity, electric force fields can be easily manipulated to cancel to zero. If you got this part wrong, you can ask about it in class. But for now, trust that the overall field pattern between the plates should look a bit like the picture at the top of the next page (only turned horizontally.) You can use the space below for brainstorming if you have questions.

Now work on the following exercise involving the Electric Field.

Suppose a parallel plate capacitor is charged so that its plates are opposite and so that the E field in the space between the plates points upward, and this E field has a strength of 200 N/Ceverywhere. The distance between plates is given.

B

4 cm E

A

  1. Figure out which plate is positive and which plate is negative. (These notes have taught you the rule for this.)
  2. Some proton has a charge of 1.6 x 10-19C. Notice the units of the E field. If I release a proton at a place between the plates, how much force will it feel?
  3. If you want to release a proton so that it crashes into one of the plates with maximum kinetic energy, at what point should you release it?
  4. If you release the proton at the point you figured out in the last question, how much work will be done on it by the time it crosses to the other plate? (These answers will be revealed as you work through the document so move quickly through educated guessing.)
  5. If the proton was released from rest, what will its kinetic energy be once it crosses to the other plate?
  6. We will care about a measurement called “Work per unit charge”. Calculate this quantity for the proton that was accelerated from point A to point B. This is easy: “per” means divide. State the units in your answer. They are J/C. You should be calculating it as 12.8 x 10-19J divided by 1.6 x 10-19C. If you didn’t get this far on your own, I was hoping that through units you would realize that a number of Newtons would come from multiplying (# N/C)(# C) = # N , because the C cancels. With the given numbers of the problem, F = (200 N/C)(1.6 x 10-19C), and this would represent the idea mentioned earlier summarized by F = qE. This would give force, and from that, energy would come from Work = F times d.
  7. Now we release a charge, Charge 2, that is negative and contains 2000 times the charge of the proton, only we release it at point B:
  • How much force acts on Charge 2?
  • How much work is done on Charge 2 as it speeds up from B to A.
  • Calculate the work per unit charge for Charge 2.

The following facts are true about the electron: it contains just as much charge as the proton. But the charge it has is opposite the type the proton has. And finally, its mass is about 1000 times less than the proton’s mass.

  1. Suppose an electron is released at point B. By the time it gets to point A, will it have the same KE, less KE, or more KE than what the proton had in question 9? State how you know.
  2. When the electron travels from one plate to the other, does it end up with the same speed as the proton, slower, or faster. State how you know.

Final notes: I hope you noticed the most important pattern in all of this:

  • E field value describes the region of space and is the same no matter which particle moves between the plates.
  • That answer you kept getting that was 8 J/C no matter whether the moving particle was the proton, Charge 2, or the electron: that 8 J/C was a Voltage. The Voltage of the plates describes the region of space and also does NOT depend on which particle is moving between the plates. For this example, the voltage was shown to be 8 J/C whether the particle was a proton, Charge 2, an electron, or whatever. The voltage of this capacitor was 8 Volts. And you can see that the definition of the volt is

Volt = Joule/Coulomb

You can also see from this that

Voltage = Work/q

So finally, understand the pattern below, and you can figure out many useful equations and actually understand them instead of mindlessly memorize them:

Forcemult by dWork

Divide Divide

byby

q q

E FieldVoltage

To move to the right from Force to Work, you multiply by distance.

To move down from Force to E field, you divide by test charge.

To move down from Work to Voltage, you also divide by test charge.

So guess what: to move to the right from E field to Voltage, you simply multiply by ___.

You add a fourth big arrow to the schematic above and label it with a math instruction.

When you move opposite the arrows, you reverse the mathematical operation. So for example, to go from Voltage to Work, you multiply by charge. To move from Voltage to E field, you divide by relevant distance. You’ll see on Page 6 if you comprehend this.

Connected Equations:

There are at least 6 equations you can figure out from the patterns. Three of them are:

(F)(d) = W takes you from upper left corner to upper right

(F)/(q) = E takes you from upper left corner to lower left

(Voltage)(q)/d = Force takes you from lower right to upper left

You figure out the three other equations that can be created by using the pattern:

Yet another way of looking at it:

I say that Work = Fd

But F = qE, so Work = qEd

But voltage, V, is Work per unit charge q

So V = Work/q

= qEd/q

So V = Ed

OR: E = V/d

This means that a common unit for the E field is Volt/meter. You need to know this, and that’s part of the reason for the Charges and Fields assignment.

The two common metric units for the E field are ( )/( ) and ( )/( ). Both of these units are identical.

In other words, 1 N/C = 1 V/m

Comprehension Test:

In the picture on Page 6, identify the point that is the midpoint between the red and blue source charges. At that point, find the numbers to plug into the math below using the contours that are NOT at zero volts, but instead using the two contours that are the closest values BOTH above zero and below zero. Why are you reading volt values off of two contours? Because you are calculating a change, and that means subtract:

(Change in Volts)/(Change in Position) = ( V)/( m) = ______

The picture should be clear enough for you to at least make an educated guess for how to fill in the missing numbers above. If you do it honestly, Page 6 will give you a good indication of your understanding. (Unless, you resort to using a tutor. Sometimes people use tutors at times like this, and it completely defeats the purpose of these notes. YOU the student are to read this paper and write what you think, UNBIASED by anybody else. Page 6 will tell you if you are right. People who immediately go to a tutor such that the tutor tells them what to do on this part are people who can never find out what they would have thought on their own. And that’s what would destroy the value of these notes. The people who try honestly will know the quality of their understanding by the end of Page 6, and they don’t need a tutor to find out. But they will need to read for meaning.)

The numerator in the calculation had to be 4.0 V, because 4.0 V is the difference between 2 V and -2 V. 2 V and -2 V define the potential values of the nearest contours that surround the spot where the orange puck is. Look at the picture.

The change in position between the 2 V and the -2 V contours is something you had to estimate from the picture by being observant. A distance can be estimated.

The red vector is something called E field. The answer to the ratio V/d calculated on the last page is also supposed to be E field.V/d = E is a Chapter 18 thing.

And from a totally different point of view (Chapter 17 instead of 18), E field is:

E = F/q = [(kqQ1)/d2 + (kqQ2)/d2]/q = (kQ1)/d2 + (kQ2)/d2

Check your understanding by plugging in the proper values for Q1, Q2, and d by being observant with the diagram above. You can get d by looking. For Q1 and Q2, I need to tell you that there is ONLY ONE red charge and ONLY ONE blue charge. From these facts, plugin the proper values of Q1, Q2, and d; calculate E. And then compare that to V/d. If the values don’t match, you’ll know that you need to refine your understanding.