More than a century after Newton, while the United States was struggling with the Articles of the Confederation, and the French were preparing for revolution, the second basic force of the universe was expressed mathematically.

It had been known for thousands of years that a piece of amber rubbed with cloth would attract bits of paper. It had been found that the force involved, the electric force, could be repulsive as well as attractive. Further, it was observed that there were two different types of charge and that one kind would cancel the other. For this reason the charges were designated "positive" and "negative" to indicate their opposite nature.

When two positive charges were brought near each other, they were found to repel as did two negative charges when they were brought close. Further, a positive and negative charge would attract each other, and the electrical force became greater with greater charge and smaller with greater distances.

In 1785 Charles Augustine de Coulomb first guessed and then experimentally demonstrated that the electric force had a form similar to the gravitational force. He found that:

F = k

q1 * q2 --------------- r2

where q₁ and q₂ are electrical charges measured in units of coulombs (to be defined at a later time), r is the distance in meters between the charges, and k is a constant experimentally determined. Coulomb found it to be

k = 9 * 10⁹ nt-m²/coul²

1. Two charges, 4 * 10^-5 C and 3 * 10^-6 C are separated by a distance of 1.4 m. What is the force between them?

F = k[(q1q2)/r2] F = (9 * 109 nt-m2/coul2)[(4 * 10-5) (3 * 10-6)/(1.4 m)2 F = .551 nt

2. The charge on an electron is 1.6 * 10^-19 C. What is the force between two electrons separated by 5 * 10^-6 cm?

F = k[(q1q2)/r2] F = (9 * 109)[(1.6 * 10-19)(1.6 * 10-19)]/ (5 * 10-8)2 F = 9.22 * 10-14 nt

3. How far apart must two 1 coulomb charges be to feel a force of 3 * 10⁴ nt? (30,000 nt is about the weight of a Cadillac.)

F = k[(q1q2)/r2] r = (kqq/F)1/2 r = [(9 * 109)(1)(1)/(3 * 104)]1/2 r = 548 m

Think about that last problem. A coulomb is a whopping charge. You will never find an isolated coulomb of charge. Furthermore, the electric force is itself an enormous force, much larger than gravitation.

The electric force is absolutely essential to the operation of the universe as we know it. In Chapter 14 we saw how gravity determines the large scale nature of the universe. It holds planets and moons in their orbits, it holds people and oceans to their planets, and holds their planets together. It determines the structure of galaxies and will eventually determine the fate of the universe as a whole. This is possible because there are enormous accumulations of mass in these objects. While G is very small, it is overwhelmed by the great mass of astronomical objects.

The small scale universe is oblivious to gravity. The tiny masses involved generate such weal gravitational fields, they may be ignored in all but the most exacting calculations. It's here that the electric force comes into its own. On a small scale electric charges are frequently not balanced, and the large value of k allows the electric force to dominate. It holds electrons in their orbits and gives atoms their structure. It bonds atoms together into molecules and holds molecules together in crystals. Because it can also be a repulsive force, it holds molecules apart, preventing objects from merging with one another. Matter as we know it could not exist without the electric force.

Let us relate the electric force to other forces we've studied in the past.

4. A charge +Q is attached to a spring of constant K as shown. Another charge, +q, is brought up so that the separation of the charges at equilibrium is d. How much is the spring compressed?

For stability, FE = FX ke(qQ/d2) = ksx x = keqQ/ksd2

5. A mass m carries a charge q and rests on a table of friction µ. How close can another charge Q be brought before the block slips?

Equilibrium implies FE = Ff k(qQ/x2) = µmg x = (kqQ/µmg)1/2

6. A mass m, carrying a charge +Q, hangs from a string. Another charge, -q is brought in at a distance s. What is the tension in the string?

At equilibrium: T = mg + k(qQ/s2)

7. A mass m with charge -q is placed directly above another equal charge, -q. At what height h will the charge levitate? (That is, float?)

m will float when: mg = k(qQ/h2) h = (kq2/mg)1/2

8. Two charges, q1 and q2 have masses m1 and m2. If they are in space separated by a distance r, a) What force will each feel? b) What acceleration will each experience?

a) Each charge will feel an identical force: FE = k(q1q2/r2)

b) a1 = f/m1 a1 = k(q1q2/r2m1) a2 = k(q1q2/r2m2)

9. A particle of mass m and charge -q orbits another particle of charge +Q at a distance r. If gravity is negligible, what is the velocity of the orbiting mass? What is its angular momentum?

FE = Fc k(Qq/r2) = mv2/r v = (kqQ/mr)1/2 Iw = mr2(v/r) Iw = mrv Iw = mr(kqQ/mr)1/2 Iw = (kQqmr)1/2

10. A bar of length L is pivoted about the center. Two charges, Q and Q are place on the ends, and two other charges, q and q are brought to a distance r as shown. What is the torque on the bar?

t = FL t = k(Qq/r2)(L/2) + k(Qq/r2)(L/2) t = k(Qq/r2)L

11. A weightless beam has a mass M at one end. A charge +Q is located part-way down the beam, and is repelled by another charge +q held at a distance x. Write an equation that expresses the equilibrium condition.

mgL1 = k(Qq/x2)L2

12. A spring holds a charge of 7 * 10^-6 C. Another charge of 3 * 10^-6 C is brought to within 8.0 cm of the first, forcing the spring to compress 3.0 cm. What is the stiffness constant of the spring? (8 cm is the final separation of the charges.)

FE = Fc ksx = kE(q1q2/r2) ks = kE(q1q2/r2x) ks = [(9 * 109)(7 * 10-6)(3 * 10-6)]/ [(.03)(.08)2] ks = 984 nt/m

13. A 450 gm mass rests on a surface of µ = 0.2. If the mass is given a 4 µC charge, how close may a 9 µC charge be brought before the mass starts to slip?

Ff = FE µmg = k(Qq/x2) x = (kQq/µmg)1/2 x = [[(9 * 109)(9 * 10-6)(4 * 10-6)]/[(.2)(.45)(9.8)]]1/2 x = .606 m

14. A 12 gm mass is tied to the floor by a string. It carries a +8 * 10^-6 C charge and is pulled up off the floor by a -4 * 10^-8 C charge 13 cm above it. What is the tension in the string?

T + mg = k(qQ/r2) T = k(Qq/r2) - mg T = (9 * 109)[(4 * 10-8)(8 * 10-6)/(.13)2] - (.012)(9.8) T = .0528 nt

15. Two 3 kg masses float in space, attracted by their mutual gravitation. What identical charge on each mass would exactly balance their attraction so that there would be no net force between them?

G(mM/r2) = k(qQ/r2) G(m2/r2) = k(q2/r2) q = m(G/k)1/2 q = (3)[((2/3) * 10-10)/(9 * 109)]1/2 q = 2.58 * 10-10 c

16. Two electrons are placed on a frictionless table at a distance of 5 * 10^-9 m. What will be the acceleration of each when they are released?

a = F/m a = [k(q2/r2)]/m a = [(9 * 109)(1.6 * 10-19)2]/ [(9.1 * 10-31)(5 * 10-9)2] a = 1.01 * 1014 m/sec2 note: that is a big acceleration!

Students who are conscientious about the signs of their answers may be disturbed at the way I have been dropping signs and winding up with only the magnitude of the force. This has been intentional. We have been dealing with problems in two dimensions using a one-dimensional, scalar notation. It's clear that strict adherence to signs would lead to negative forces being attractive and positive being repulsive, but this wouldn't solve the problem. Look at this situation:

Each calculation of a force will be negative, but what insight will that give as to the direction of the force? None, unless we adopt a vector notation that clearly states the 2-dimensional location of one particle relative to the other. You will learn such a notation in more advanced courses, but at this stage it will be more valuable to simply look at the situation, think about it, and set up your equations appropriately.

17. In the hydrogen atom an electron orbits a proton at a distance of 5.3 * 10^-11 m. What orbital velocity must it have?

We know the charge on a proton and electron are 1.6 * 10-19 coul. The mass of an electron is 9.1 * 10-31 kg. In orbit: k(qQ/r2) = mv2/r v = (kqq/mr)1/2 v = [[(9 * 109)(1.6 * 10-19)2]/ [(9.1 * 10-31)(5.3 * 10-11)]] v = 2.19 * 106 m/sec

18. A +4.6 * 10^-7 coul charge and a +2.1 * 10^-6 C charge are tethered together by a 5.3 cm string. What is the tension in the string?

For each charge the force of repulsion is balanced by the tension in the string. T = k(qQ/r2) T = (9 * 109)[(4.6 * 10-7)(2.1 * 10-6)/(.053)2] T = 3.10 nt

19. Three charges are held together with threads as shown. What is the tension in the left-hand thread? (Hint: first draw a diagram of the forces on the 4 * 10^-4 charge.)

T = F1 + F2 T = (9 * 109)[(4 * 10-4)(2 * 10-4)/(3)2] + (9 * 109)[(4 * 10-4)(6 * 10-4)/(7)2] T = 124 nt

20. Three charges are tethered as shown. What is the tension in the right-hand string?

Isolate forces on the 12µC charge: T + F1 = F2 T = F2 - F1 T = (9 * 109)[(18 * 10-6)(12 * 10-6)/(.29)2] - (9 * 109)[(2 * 10-6)(12 * 10-6)/(.11)2] T = 5.26 nt

21. Three charges are positioned as shown. What is the net force on the 3 * 10^-4 C charge, and what is its direction?

The charge feels two forces: F1 = (9 * 109)[(3 * 10-4)(2 * 10-5)/(.6)2] F1 = 150 nt F2 = (9 * 109)[(3 * 10-4)(5 * 10-5)/(.8)2] F2 = 211 nt F = (F12 + F22)1/2 F = 259 nt Ø = tan-1(150/211) Ø = 35.4°

22. Again, find the magnitude and direction of the force on the 7 µC charge.

F1 = (9 * 109)[(7 * 10-6)(80 * 10-6)/(.08)2] F1 = 788 nt F2 = (9 * 109)[(7 * 10-6)(30 * 10-6)/(.10)2] F2 = 189 nt F = (F12 + F22)1/2 F = 810 nt Ø = tan-1(788/189) Ø = 76.5°

23. Two small spheres of mass m are suspended by weightless threads of length L as illustrated. When given equal charges, the spheres separate, forming equal angles q with the vertical. Given m, g, L, k, and q, find:
a) the value of q.
b) Compare the angle on the left with that on the right if the charge on the left is doubled. (is it smaller, larger, same?)
c) Compare the angles on left and right if the mass on the left is doubled.

a) The string will adjust itself up or down until the following condition is reached: tanØ = [k(qq/r2)]/[mg] tanØ = kq2/mgr2 q = r(mgtanØ/k)1/2 q = 2LsinØ(mgtanØ/k)1/2

b) If one charge is doubled, the force on both charges is doubled. Symmetry is still present and the angles will be equal.

c) if the left mass is larger, the left angle will be smaller.

24. Find the magnitude and direction of the force on the 4 * 10^-6 C charge.

F1 = (9 * 109)[(2 * 10-6)(4 * 10-6)/(.12)2] F1 = 5.00 nt F2 = (9 * 109)[(5 * 10-6)(4 * 10-6)/ (.122 + .092)] F2 = 8.00 nt Ø = tan-1(9/12) Ø = 36.9° F2 can be broken into vertical and horizontal components: F2x = F2cosØ2 F2x = (8.00)cos(36.9) F2x = 6.4 nt F2y = F2sinØ F2y = (8.00)sin(36.9) F2y = 4.8 nt F = (1.42 + 4.82)1/2 F = 5.0 nt Ø = tan-1(4.8/1.4) Ø = 73.7°

25. One last vector problem. Find the magnitude and direction on the 7 * 10^-7 C charge.

F1 = (9 * 109)[(2 * 10-6)(7 * 10-7)/(.17)2] F1 = .436 nt F2 = (9 * 109)[(9 * 10-7)(7 * 10-7)/ (.232 + .172)] F2 = .0693 nt F2x = F2cosØ F2x = (.0693)cos(36.5) F2x = .0557 nt F2y = F2sinØ F2y = (.0693)sin(36.5) F2y = .0412 nt Net vertical force = force up - force down: Net vertical force = .436 nt -.0412 Net vertical force = .395 nt F = (.3952 + .05572)1/2 F = .399 nt Ø = tan-1(.395/.0557) Ø = 82.0°