Electric Charge, Coulomb's Law, and Electric Fields: A Comprehensive Introduction, Study notes of Physics

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Worked Examples from Introductory Physics

(Algebra–Based)

Vol. IV: Electricity

David Murdock, TTU

January 20, 2008

• 1 Electric Charge and Coulomb’s Law
  • 1.1 The Important Stuff
    • 1.1.1 Electric Charge
    • 1.1.2 Some Facts About Electric Charge
    • 1.1.3 Coulomb’s Law
  • 1.2 Worked Examples
    • 1.2.1 Electric Charge
    • 1.2.2 Coulomb’s Law
• 2 The Electric Field
  • 2.1 The Important Stuff
    • 2.1.1 The Electric Field
    • 2.1.2 Finding the Electric Field
    • 2.1.3 Continuous Distributions; Sheets of Charge
    • 2.1.4 Electric Field Lines
    • 2.1.5 Conductors
  • 2.2 Worked Examples
    • 2.2.1 The Electric Field
    • 2.2.2 Finding the Electric Field
• 3 Electric Potential Energy; Electric Potential
  • 3.1 The Important Stuff
    • 3.1.1 Electric Potential Energy and Electric Potential
    • 3.1.2 Calculating the Electric Potential
    • 3.1.3 Equipotentials; Relation Between E and V
    • 3.1.4 Capacitance
    • 3.1.5 Dielectrics
    • 3.1.6 Capacitors and Energy
  • 3.2 Worked Examples
    • 3.2.1 Electric Potential Energy and Electric Potential
    • 3.2.2 Calculating the Electric Potential 4 CONTENTS
    • 3.2.3 Capacitance
    • 3.2.4 Dielectrics
    • 3.2.5 Capacitors and Energy
• 4 Electric Current and Resistance
  • 4.1 The Important Stuff
    • 4.1.1 Electric Current
    • 4.1.2 Ohm’s Law
    • 4.1.3 Resistance and Resistivity
    • 4.1.4 Electric Power
    • 4.1.5 Series and Parallel Circuits
    • 4.1.6 Kirchhoff’s Rules
  • 4.2 Worked Examples
    • 4.2.1 Electric Current
    • 4.2.2 Ohm’s Law
    • 4.2.3 Resistance and Resistivity
    • 4.2.4 Electric Power
    • 4.2.5 Resistors in Series and in Parallel

Chapter 1

Electric Charge and Coulomb’s Law

1.1 The Important Stuff

1.1.1 Electric Charge

In the latter part of the 18th century it was realized that any sample of matter has a property which is as fundamental as its mass. This property is the electric charge of the sample. Electric charge can be detected because it gives rise to electric forces. The reason that we don’t see electric phenomena more often than we do is that electric charges come in two types —positive and negative— and usually the two types occur in equal numbers so that they add to give zero net charge. But when we can separate positive and negative charges we observe electric forces on a large scale. In the SI system, electric charge is measured in Coulombs. Throughout our study of electromagnetism we will derive other electrical units based on the Coulomb and the units already encountered in mechanics. After decades of study of the electrical properties of matter, it was found that the fun- damental charges in nature occur in integer multiples of the elementary charge e,

e = 1. 602 × 10 −^19 C (1.1)

In discussing this property of charge we often say that electric charge is quantized. In the atom, the nucleus has a charge which is a multiple of +e while the orbiting electrons each have a charge of −e. The charge of the nucleus comes from the constituent protons, each of which has a charge of +e; the neutrons in the nucleus have no charge.

1.1.2 Some Facts About Electric Charge

Electric charges can be separated by rubbing, as when you rub a plastic rod with some roadkill; see Fig. 1.1. Then one of the objects will obtain a positive charge and the other

1

1.2. WORKED EXAMPLES 3

charges; finally, it gets weaker as the distance between the charges increases. But the force is not inversely proportional to the distance, it is inversely proportional to the square of the distance. The law for the magnitude of the electric force between two small charges q 1 and q 2 separated by a distance r is

F = k

|q 1 q 2 | r^2

where k = 8. 99 × 10 9 N C·m 2 2 (1.2)

This is usually called Coulomb’s law. The constant k will come up often in our examples but later on it will be easier to work with the constant  0 , which is related to k by

k =

4 π 0

so that  0 has the value

 0 =

4 πk

= 8. 85 × 10 −^12 C

2 N·m^2 (1.3)

The electric force given by Coulomb’s law is similar to Newton’s law for the gravitational force (from first semeseter) in that both are inverse–square laws; the force is inversely proportional to the square of the distance between the particles.

If we plug some easy numbers into Eq. 1.2 we find that if two 1.0 C charges are separated by a meter, then each one experiences a repulsive force of about 9. 0 × 109 N, which is an enormous force. In this sense, 1 C is a huge amount of charge; typically the charges which one would encounter in real life are of the order of μC (10−^6 C) or nC (10−^9 C).

When a charge Q is in the vicinity of several other charges (q 1 , q 2 , etc.) the net force on Q is found by adding up the individual forces from the other charges. Of course, this is a vector sum of the forces.

1.2 Worked Examples

1.2.1 Electric Charge

1. How many electrons must you have to get a total charge of − 1 .0 C? How many moles of electrons is this?

4 CHAPTER 1. ELECTRIC CHARGE AND COULOMB’S LAW

Since each electron has a charge of − 1. 6 × 10 −^19 C, the number of electrons required is

N =

(− 1 .0 C)

(− 1. 6 × 10 −^19 C)

= 6. 2 × 1018

A mole of any kind of particle is NAvo = 6. 02 × 1023 (Avogadro’s number) of those particles. Here we have 6. 2 × 1018 electrons and that is

n =

N

NAvo

(6. 2 × 1018 )

(6. 02 × 1023 )

= 1. 04 × 10 −^5 moles

2. A metal sphere has a charge of +8. 0 μC. What is the net charge after 6. 0 × 1013 electrons have been placed on it? [CJ6 15-2]

The total charge of 6. 0 × 1013 electrons is

Qelec = (6. 0 × 1013 )(−e) = (6. 0 × 1013 )(− 1. 60 × 10 −^19 C) = − 9. 6 × 10 −^6 C = − 9. 6 μC

After this charge has been added to the metal sphere its total charge is

Qsph = +8. 0 μC − 9. 6 μC = − 1. 6 μC

1.2.2 Coulomb’s Law

3. A charge of 4. 5 × 10 −^9 C is located 3 .2 m from a charge of − 2. 8 × 10 −^9 C. Find the electrostatic force exerted by one charge on another. [SF7 15-1]

This will be a force of attraction between the two charges since they are of opposite signs. The magnitude of this force is given by Coulomb’s law, Eq. 1.2,

F = k

|q 1 q 2 | r^2 = (8. 99 × 10 9 N C·m 2 2 )

(4. 5 × 10 −^9 C)(2. 8 × 10 −^9 C)

(3.2 m)^2

= 1. 1 × 10 −^8 N

The charges will attract one another with a force of magnitude 1. 1 × 10 −^8 N.

4. An alpha particle (charge=+2. 0 e) is sent at high speed toward a gold nucleus (charge=+79e). What is the electrical force acting on the alpha particle when it is 2. 0 × 10 −^14 m from the gold nucleus? [SF7 15-3]

6 CHAPTER 1. ELECTRIC CHARGE AND COULOMB’S LAW

(a)

12 x 10-9^ C -18 x 10-9^ C -3.0 x 10-9^ C^ -3.0 x 10-9^ C

(b)

Figure 1.3: (a) Conducting spheres are given different charges. (b) Charges on the spheres after being joined by a conducting wire.

5.00 nC 6.00 nC

-3.00 nC

0.300 m

0.100 m

x

y

Figure 1.4: Charges in Example 6

direction of the electrostatic force on the charge at the origin. [SF7 15-11]

Let’s call the 6.00 nC charge q 1 and the − 3 .00 nC charge q 2. (The charge at the origin is Q = +5.00 nC.) The force from q 1 is repulsive and points to the right. The force from q 2 is attractive and points downward, as shown in Fig. 1.5. We need to find the magnitudes of F 1 and F 2 and then add those two force vectors. From Coulomb’s law we get the magnitude of F 1 ; since charge q 1 is at a distance r 1 =

5.00 nC

Q

F 1

F 2

F net

Figure 1.5: Forces on Q in Example 6

1.2. WORKED EXAMPLES 7

0 .300 m from Q,

F 1 = k

|Qq 1 | r^21

= (8. 99 × 10 9 N·m

2 C^2 )

(5. 00 × 10 −^9 C)(6. 00 × 10 −^9 C)

(0.300 m)^2

= 3. 00 × 10 −^6 N

Likewise, the magnitude of F 2 is

F 2 = k

|Qq 2 | r^22

= (8. 99 × 10 9 N C·m 2 2 )

(5. 00 × 10 −^9 C)(3. 00 × 10 −^9 C)

(0.100 m)^2

= 1. 35 × 10 −^5 N

Then the total force on Q has the components

Fx = − 3. 00 × 10 −^6 N Fy = − 1. 35 × 10 −^5 N

What is the magnitude and direction of this vector? Its magnitude is

F =

√ (− 3. 00 × 10 −^6 N)^2 + (− 1. 35 × 10 −^5 N)^2 = 1. 38 × 10 −^5 N

and the direction we can find from

θ = tan−^1

(− 1. 35 × 10 −^5 )

(− 3. 00 × 10 −^6 )

= 77. 5 ◦^ − 180 ◦^ = − 103 ◦

(Note, we subtract 180◦^ from the simple answer because the direction of the force is in the third quadrant.) The net force on Q has magnitude 1. 38 × 10 −^5 N and points at an angle of − 103 ◦^ from the +x axis.

7. Two small metallic spheres, each of mass 0 .20 g are suspended as pendulums by light strings from a common point as shown in Fig. 1.6. The spheres are given the same electric charge and it is found that they come to equilibrium when each string is at an angle of 5. 0 ◦^ with the vertical. If each string is 30 .0 cm long, what is the magnitude of the charge on each sphere? [SF7 15-15]

From simple trig we can calculate the distance between the two spheres. If this distance is x, then x = 2(30.0 m) sin 5. 0 ◦^ = 5.23 cm = 5. 23 × 10 −^2 m

1.2. WORKED EXAMPLES 9

Now consider the forces acting on one of the spheres, say the one on the right. These are shown in Fig. 1.7, where we also note (for reference) the location of other sphere. The right sphere experiences a force of electric repulsion from the left sphere. The forces are the force of gravity (mg, downward), the tension of the string (magnitude T ; it pulls at an angle 5. 0 ◦ from the vertical) and the electric repulsive force. From Coulomb’s law, the magnitude of the latter is

Felec = k

q^2 x^2 where q is the magnitude of the charge on each sphere. The sphere is in equilibrium, so the forces must sum to zero. The vertical forces cancel out, giving us:

T cos 5. 0 ◦^ = mg =⇒ T = mg cos 5. 0 ◦^

(2. 00 × 10 −^4 kg)(9. 80 m s 2 ) cos 5. 0 ◦^

= 1. 97 × 10 −^3 N

The horizontal forces cancel out and this gives:

T sin 5. 0 ◦^ = Felec = k

q^2 x^2

which lets us solve for q:

q^2 =

T x^2 sin 5. 0 ◦ k

(1. 97 × 10 −^3 N)(5. 23 × 10 −^2 )^2 sin 5. 0 ◦ (8. 99 × 10 9 N·m

2 C^2 )^

= 5. 22 × 10 −^17 C^2

so then q = 7. 2 × 10 −^9 C = 7.2 nC

10 CHAPTER 1. ELECTRIC CHARGE AND COULOMB’S LAW

12 CHAPTER 2. THE ELECTRIC FIELD

q

E

q

E P (^) P

r

(a) (b)

Figure 2.2: (a) Point P is a distance r away from charge q. If q is positive, the electric field points away from q. (b) If q is negative, the electric field points toward q. In both cases the magnitude of E is given by E = k|q|/r^2.

When we use this equation we mean that after we put Q in place all the little charges q 1 , q 2... are in the same places they were when we deduced the value of E from their values and positions! This will be true in practice if the “test charge” Q is small. Thus we give a practical definition of the E field as

E =

F

Q

where Q is a small charge (2.2)

From Eq. 2.2 we see that the electric field is a vector and has units of N/C.

We note that finding the electric field is more useful than finding the force on a specific charge since once we have the E field we simply multiply by the charge Q to get the force, as given by Eq. 2.1.

2.1.2 Finding the Electric Field

It follows from Coulomb’s law that at a point which is a distance r from a point charge q, the magnitude of the electric field is

Ept−ch = k

|q| r^2

and the direction of the field is away from q if q is positive and toward q if q is negative. This is shown in Fig. 2.2.

When we need to find the electric field due a collection of point charges we find the electric field due to each charge and then find the (vector) sum.

2.1. THE IMPORTANT STUFF 13

P

E

z

Figure 2.3: Point P is at some distance z above an infinte plane of charge with charge density σ. If σ is positive the E field points away from the sheet and has magnitude σ/(2 0 ).

2.1.3 Continuous Distributions; Sheets of Charge

Many charged objects we encounter are not sets of points charges; rather they are continuous distributions of charge. If a two-dimensional region of space contains a charge we can talk about its charge per unit area, or surface charge density. Surface charge density is usually given the symbol σ; it has units of C/m^2. The simplest case of a surface charge is that of an infinite planar sheet of charge with uniform charge density σ. We want to know the value of the electric field E at a point P which is a distance z from the plane; see Fig. 2.3. It turns out that the answer does not depend on z. If σ is positive, the electric field at P points away from the sheet and has magnitude

Einf−sh =

σ 2  0

Here it is easiest to express the result using the constant  0 introduced in Eq. 1. If the sheet has a negative charge density then the field points toward the sheet and the magnitude of the field is E = |σ|/(2 0 ).

Next we take the case of the two very large, flat, parallel sheets of charge, as shown in Fig. 2.4. A total charge of +Q has been placed on one sheet and a charge of −Q on the other. We assumed the charge is spread around uniformly so that the charge density of the positively-charge sheet is σ = Q A. This situation can arise when equal and opposite charges are placed on metal plates which are held apart at some distance. (Such a device is called a parallel–plate capacitor.) Our approximation is suitable for the case where the plates are separated by a distance which is small compared with the linear size of the plates. From Eq. 2.4 it follows that the magnitude of the E field between the plates is twice that of the single sheet,

Einf−sh =

σ  0

Q

 0 A

2.1. THE IMPORTANT STUFF 15

+q +q

(a) (b)

Figure 2.5: (a) A representation of the electric field around a point charge using individual vectors. (b) Representation of the electric field around a point charge using field lines.

+q (^) -q

Figure 2.6: Field lines of an electric dipole.

16 CHAPTER 2. THE ELECTRIC FIELD

F elec = qE

F grav = mg

q

m

Figure 2.7: Forces on the charged mass in Example 1. The electric force is upward (in the same direction as the E field). The force of gravity is downward.

But the mathematics of the electric force tell us that the number of field lines that we draw originating on a charge should be proportional the the size of the charge. If we follow that rule, then the magnitude of the electric field can be judged from the density of field lines at any point. If the lines are closely space, the electric field is strong at that place.

2.1.5 Conductors

In conductors any excess charge is free to move through the material.

2.2 Worked Examples

2.2.1 The Electric Field

1. An object with a net charge of 24 μC is placed in a uniform electric field of 610 N C , directed vertically. What is the mass of the object if it “floats” in the electric field? [SF7 15-17]

The forces acting on this object (of mass m and charge q are shown in Fig. 2.7. The force of gravity has magnitude mg and points downward. The electric force, from Eq. 2. has magnitude qE and points upward. (Here the charge q is positive so that the force points in the same direction as the E field.) The object ”floats” so the net force on it must be zero. Hence:

qE = mg =⇒ m =

qE g