Coulomb established the law describing the force between charged particles and viewed it much as Newton did gravity action at a distance an inexplicable force between separated objects. In the 19th century Michael Faraday did a superb series of experiments that clearly delineated all the fundamental principles that were then known about electricity and magnetism. In doing this, Faraday developed a somewhat different view of electrostatic attraction. He believed that the first charged particle conditioned the space around it, and the second charge responded to that condition. This alteration in space was given the name electric field. That is what we'll discuss in this chapter.

First some warm-up problems to get the idea:

1. Various charges are to be placed 10 cm away from a 5 * 10^-6 coulomb charge. In each case, determine the force acting on the second charge.
a) 4 * 10^-6 C
b) 2 * 10^-6 C
c) 8 * 10^-6 C

a) F = k(qq/r2) F = (9 * 109)[(5 * 10-6)(4 * 10-6)/(.12)] F = 18.0 nt

b) F = k(qq/r2) F = (9 * 109)[(5 * 10-6)(2 * 10-6)/(.12)] F = 9.0 nt

c) F = k(qq/r2) F = (9 * 109)[(5 * 10-6)(8 * 10-6)/(.12)] F = 36 nt

2. Rosa Corn pulls an 8 * 10^-4 coulomb charge out of her purse and holds it in front of her. She feels a force of 20 nt on it.
a) What force will she feel when she holds a 4 * 10^-4 C charge?
b) What force when she holds a 40 * 10^-4 C charge?
c) What force when she holds a charge q?

a) F = 1/2 * 20 F = 10 nt

b) F = 5 * 20 F = 100 nt

c) [q/(8 * 10-4)] * 20 nt

3. A 2 µC charge is placed near another charge and feels a force of 3 newtons.
a) What charge in the same place would feel a force of 6 newtons?
b) What charge would feel a force of 4 newtons?

a) To feel twice the force we need twice the charge. q = (2µC)(6nt/3nt) q = 4µC

b) q = (2µC)(4nt/3nt) q = 2.7µC

Look back over problems 1-3. If you are unsure of yourself, review them. The key concept is that under fixed conditions, the force on a charged particle will be proportional to its charge. Mathematically we may say

F «» q or F = kq

Our notation will be F = qE where E is the electric field strength. We think of electric field as the force per unit charge, or

E = F/q

Its units are newtons/coulomb in the SI system.

The quick way to find if there is an electric field present is to whip a coulomb out of your pocket and see if there is a force on it. (In practice we use very small charges, so they will not disturb the existing charge structure in the room.)

4. A 4 * 10^-3 C charge feels a force of 14 nt. What is the electric field in its vicinity?

E = F/q E = (14 nt)/(4 * 10-3C) E = 3.5 * 103 nt/C

5. A 0.0062 C charge is placed in a field of 7 * 10³ nt/C. What force will be exerted on it?

E = qE E = (.0062 C)(7 * 103 nt/C) E = 43.4 nt

6. In a 5.6 * 10⁵ nt/C field, a charge feels a 41 nt force. What is the size of the charge?

E = qE q = F/E q = (41 nt)/(5.6 * 105 nt/C q = 7.3 * 10-5C

7. In a particular region of space a 6 * 10^-5 C charge feels a force of 11 nt. What force will be felt by a 3 * 10^-5 C charge?

One way: E = F/q1 E = (11 nt)/(6 * 10-5C) E = 1.83 * 105 nt/C F = q2E F = (3 * 10-5)(1.83 * 105) F = 5.5 nt Another way: F «» q 1/2 the force is 5.5 nt

8. An object masses 1.6 gm and is charged with 4.7 * 10^-6 C. What electric field would keep it from falling to Earth?

qE = mg E = mg/q E = [(1.6 * 10-3kg)(9.8 m/sec2)/(4.7 * 10-6C)] E = 3.3 * 103 nt/C

9. A mass m rests on a frictional surface µ. If the mass carries a charge q, what horizontal electric field will be required to make the mass slide?

Ff = FE µmg = qE E = µmg/q

10. How fast will an electron accelerate in a 750 nt/C electric field?

a = F/m a = qE/m a = [(1.6 * 10-19)(750)]/(9.1 * 10-31) a = 1.32 * 1014 m/sec2

11. A 3 * 10^-5 C charge and a 7 * 10^-6 C charge are separated by 0.8 meters.
a) What force is felt by the 7 * 10^-6 C charge?
b) What electric field does it feel? (That is, what is the force per unit charge?)

a) F = k(qq/r2) F = (9 * 109)[(3 * 10-5)(7 * 10-6)/(.82)] F = 2.95 nt

b) E = F/q E = (2.95 nt)/(7 * 10-6coul) E = 4.21 * 105 nt/C

12. Two charges, Q and q are separated by a distance R.
a) What force is exerted on charge q?
b) What electric field does charge q feel?

a) F = k(qQ/R2)

b) E = F/q E = k(qQ/R2)/q E = k(Q/R2)

The equation derived in the last question is an important one. The electric field from a point charge Q, is given by

E = k(Q/R²)

13. What is the electric field 15 cm from a 6 µC charge?

E = k(Q/R2) E = (9 * 109)[(6 * 10-6)/(.152)] E = 2.40 * 106 nt/C

Let's break for a minute to discuss some common errors. First note that we now have two equations for the electric force. One is Coulomb's law:

F = k(q₁q₂/r²)

The other is the definition of electric field:

F = qE

Coulomb's law applies only to forces between point charges. In more advanced courses you will see how any charge distribution may be broken up into point charges and treated with Coulomb's law. In this class it will be used only with a small number of distinct point charges.

The electric field equation applies in all other circumstances. For some mysterious reason students frequently apply the wrong equation. Keep in mind the two equations and what they say, and you should have no problem.

14. What is the force on a 3 * 10^-7 C charge in a 5000 nt/C field?

F = qE F = (3 * 10-7coul)(5 * 103 nt/coul) F = 1.50 * 10-3 nt

15. What is the force on a 3 * 10^-7 C charge positioned 11 cm from a 5 * 10^-4 C charge?

F = k(qq/r2) F = (9 * 109)[(3 * 10-7)(5 * 10-4)/(.112)] F = 112 nt

16. What is the acceleration of an electron 0.9 nm from a proton?

a = F/m a = [k(qq/r2)]/m a = [(9 * 109)[(1.6 * 10-19)(1.6 * 10-19)]/ (.9 * 10-92)] a = 3.13 * 1020 m/sec2

17. What is the acceleration of a 2 gm mass with a 38 µC charge, in a 7.4 * 10⁴ nt/C electric field?

a = F/m a = qE/m a = [(38 * 10-6)(7.4 * 104)]/(2 * 10-3) a = 1.41 * 103 m/sec2

18. Two flat metal plates are separated by 8 cm. A uniform electric field of 3 * 10⁴ nt/C exists between the plates. How fast will an electron, released at the negative plate, be traveling when it strikes the positive plate?

First, the acceleration: a = F/m a = qE/m a = [(1.6 * 10-19)(3 * 104)]/(9.1 * 10-31) a = 5.27 * 1015 m/sec2 Then the velocity: v2 = 2as v2 = 2(5.27 * 1015)(.08) v2 = 8.44 * 1014 m2/s2 v = 2.90 * 107 m/sec

19. A 4 gm ball hangs in a gravitational field and is pulled sideways by an electric field. If the ball carries 40 µC and the string hangs at 27°, what is the strength of the E-field?

For equilibrium, the resultant force must pull along the direction of the string. tan27 = qE/mg tan27 = [(40 * 10-6)E]/[(.004)(9.8)] E = 499 nt/C

20. What is the electric field at point P between two positive charges as shown?

Two electric fields are generated: E1 = kq/r12 E1 = [(9 * 109)(3 * 10-5)]/(22) E1 = 6.75 * 104 E2 = kq/r22 E2 = [(9 * 109)(7 * 10-6)]/(32) E2 = 7.00 * 103 E = E1 - E2 E = (67.5 - 7) * 103 E = 60.5 * 103 nt/C

21. What is the electric field at points A and B?

Using the logic of #20: EA = E1 + E2 EA = (9 * 109)[(5 * 10-6)/(22) + (3 * 10-6)/(32)] EA = 1.43 * 104 nt/C EB = E1 - E2 EB = (9 * 109)[(5 * 10-6)/(82) - (3 * 10-6)/(32)] EB = -2.30 * 103 nt/C

A pictorial representation of an electric field is often helpful. We have calculated the magnitude of an E-field, now we get the direction by seeing which way a small positive charge will be forced by the field.

22. On a piece of paper designate a positive point charge. In the space around it find the magnitude and direction of the electric field. (For magnitude, simply identify a stronger field with a longer arrow.)

PLACE IMAGES HERE: IMAGE IMAGE Note that the positive test charges are forced away from the positive point charge. The closer to q a positive charge is, the larger the repelling force. Combining the results gives this picture of the field'

23. Repeat problem 22 using a negative point charge.

PLACE IMAGES HERE: IMAGE IMAGE Note that when a point is too close to the charge, the picture of its E-vector goes right through the charge! Don't be alarmed, the length of the vector merely tells the strength of E at the point.

In the next few problems you will need a ruler. Trace the problem onto another paper and find the electric fields by actually measuring the distance from charge to point and drawing in the E-field vectors on a scale of 1 cm = 1 * 10⁶ nt/C.

24. A +6 * 10^-7 C charge is located as shown. Find and diagram the E-fields at the points shown. (Remember, a 1 cm long arrow represents 1 * 10⁶ nt/C.)

Distances between charge and points: rA = 4.2 cm rB = 1.5 cm rC = 8.5 cm EA = (9 * 109)[(6 * 10-7)/(.0422)] EA = 3.1 * 106 nt/C 3.1 cm vector EB = (9 * 109)[(6 * 10-7)/(.0152)] EB = 2.4 * 107 nt/C 24 cm vector EC = (9 * 109)[(6 * 10-7)/(.0852)] EC = 7.4 * 105 nt/C .74 cm vector IMAGE

25. Diagram the E-field vectors at the points indicated.

EA = (9 * 109)[(2 * 10-7)/(.0582)] EA = 5.4 * 105 nt/C EB = (9 * 109)[(2 * 10-7)/(1.0212)] EB = 4.1 * 106 nt/C IMAGE

26. The field at point P is the result of two E-fields. Find and diagram each E-field at point P, then diagram their vector sum.

E1 = (9 * 109)[(5 * 10-7)/(.052)] E1 = 1.8 * 106 nt/C E2 = (9 * 109)[(4 * 10-7)/(.032)] E2 = 4 * 106 nt/C

27. Find the component fields and the resultant fields at the points shown. (Continue to use the 1 cm = 1 * 10⁶ nt/c scale.)

IMAGE

If you think of the electric field diagram as showing the direction of the field vector at every point in space, you can sketch out a field without doing calculations and be reasonably correct.

28. Sketch the field around an isolated positive point charge.

IMAGE

29. Sketch the field around an isolated negative point charge.

IMAGE

30. Sketch the field around an electric dipole that is, two equal but opposite charges separated by some distance.

By considering the field point-by-point, we come to this. IMAGE

31. Sketch the field around two equal positive point charges separated by some distance.

By the same point-by-point method, the following field is derived: IMAGE

E-field diagrams tell you more than the direction of the electric force. By looking at the concentration of field lines, you can get an idea of the strength of the field. Look at the last few problems. Note that near the charges the lines are concentrated far from them, they are dispersed.

32. Two large, flat plates are separated by a small distance and are oppositely charged. Find the direction of the field at points A, B, and C. Sketch the total field.

IMAGE IMAGE Except near the edges, teh field is directly away from the positive and directly toward the negative. This results in a very uniform field.

Parallel plates are important in physics, because they can generate uniform electric fields. Note that the direction and concentration of lines is unchanged from top to bottom and side to side.

33. One large, flat plate is positively charged. Find the direction of the field at points near its surface. Sketch the total field.

For most students this seems a surprising result shouldn't the field be more intense close to the plate? As long as you stay close to the plate, the direction of the field is always directly away from the plate that's easy to see. But parallel field lines implies the field strength does not vary with distance, and that is a bit hard to swallow. Think of it this way as our positive test charge moves higher above the large late, it ";sees"; a greater area, and hence a greater charge. The increased effective charge exactly counters the increased distance.

The pictures show circles of charge that most influence the test charge at various heights. Charge outside the circles will pull approximately parallel to the surface and contribute very little to the total force.

There are a couple of things to keep in mind when you're sketching field diagrams.

1) Near a point charge the field radiates uniformly outward in all directions.
2) Near the surface of a plate the field is always perpendicular to the plate surface.
3) Field lines NEVER cross.

34. Sketch the fields associated with the following:

a)

b)

We will not consider more complex cases, but if all this makes sense to you, you might want to try some charge distributions where the positive and negative charges do not balance.

35. Sketch the fields on a separate sheet of paper:

a)

b)

c)