1. Chapter 21 Electric Charge and Electric Field
Chapter 21 opener. This comb has acquired a static electric charge, either from passing through hair, or being rubbed by a cloth or paper towel. The electrical charge on the comb induces a polarization (separation of charge) in scraps of paper, and thus attracts them. Our introduction to electricity in this Chapter covers conductors and insulators, and Coulomb’s law which relates the force between two point charges as a function of their distance apart. We also introduce the powerful concept of electric field. Electrostatic Bell Demo

2. 21-5 Coulomb’s Law
Conceptual Example 21-1: Which charge exerts the greater force?
Two positive point charges, Q1 = 50 μC and Q2 = 1 μC, are separated by a distance . Which is larger in magnitude, the force that Q1 exerts on Q2 or the force that Q2 exerts on Q1?
Solution: Writing down Coulomb’s law for the two forces shows they are identical. Newton’s third law tells us the same thing.

3. 21-5 Coulomb’s Law Example 21-2: Three charges in a line.
Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges.
Solution: Coulomb’s law gives the magnitude of the forces on particle 3 from particle 1 and from particle 2. The directions of the forces can be found from the geometrical arrangement of the charges (NOT by putting signs on the charges in Coulomb’s law, which is what the students will want to do). F = -1.5 N (to the left).

4. ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 3R +Q
+4Q R 2R 1 2 3 4 5
Answer: 2

5. ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 3R +Q
+4Q R 2R 1 2 3 4 5
The force on Q0 due to +Q is: F = k(Q0)(Q)/R2
The force on Q0 due to +4Q is: F = k(Q0)(4Q)/(2R)2
Since +4Q is 4 times bigger than +Q, Q0 needs to be farther from +4Q. In fact, Q0 must be twice as far from +4Q, since the distance is squared in Coulomb’s law.

6. ConcepTest 21.5c Proton and Electron III
A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
Answer: 3 p e

7. ConcepTest 21.5c Proton and Electron III
A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
By Newton’s 3rd law, the electron and proton feel the same force. But, since F = ma, and since the proton’s mass is much greater, the proton’s acceleration will be much smaller!
Thus, they will meet closer to the proton’s original position. p e
Follow-up: Which particle will be moving faster when they meet?

8. 21-5 Coulomb’s Law
Example 21-3: Electric force using vector components.
Calculate the net electrostatic force on charge Q3 shown in the figure due to the charges Q1 and Q2.
Figure Determining the forces for Example 21–3. (a) The directions of the individual forces are as shown because F32 is repulsive (the force on Q3 is in the direction away from Q2 because Q3 and Q2 are both positive) whereas F31 is attractive (Q3 and Q1 have opposite signs), so F31 points toward Q1. (b) Adding F32 to F31 to obtain the net force.
Solution: The forces, components, and signs are as shown in the figure. Result: The magnitude of the force is 290 N, at an angle of 65° to the x axis.

9. 21-5 Coulomb’s Law Conceptual Example 21-4: Make the force on Q3 zero.
In the figure, where could you place a fourth charge, Q4 = -50 μC, so that the net force on Q3 would be zero?
Solution: The force on Q3 due to Q4 must exactly cancel the net force on Q3 from Q1 and Q2. Therefore, the force must equal 290 N and be directed opposite to the net force calculated in the previous example.

10. 21-6 The Electric Field
The electric field is defined as the force on a small charge, divided by the magnitude of the charge:
Figure Force exerted by charge Q on a small test charge, q, placed at points A, B, and C.

11. 21-6 The Electric Field An electric field surrounds every charge.
Figure An electric field surrounds every charge. P is an arbitrary point.

12. 21-6 The Electric Field
For a point charge:

13. 21-6 The Electric Field Force on a point charge in an electric field:
Figure (a) Electric field at a given point in space. (b) Force on a positive charge at that point. (c) Force on a negative charge at that point.

14. 21-6 The Electric Field Example 21-5: Photocopy machine.
A photocopy machine works by arranging positive charges (in the pattern to be copied) on the surface of a drum, then gently sprinkling negatively charged dry toner (ink) particles onto the drum. The toner particles temporarily stick to the pattern on the drum and are later transferred to paper and “melted” to produce the copy. Suppose each toner particle has a mass of 9.0 x kg and carries an average of 20 extra electrons to provide an electric charge. Assuming that the electric force on a toner particle must exceed twice its weight in order to ensure sufficient attraction, compute the required electric field strength near the surface of the drum.
Solution: The electric force on each particle must be at least twice its weight; this gives E = 5.5 x 103 N/C.

15. 21-6 The Electric Field
Example 21-6: Electric field of a single point charge.
Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C.
Figure Example 21–6. Electric field at point P (a) due to a negative charge Q, and (b) due to a positive charge Q, each 30 cm from P.
Solution: Substitution gives E = 3.0 x 105 N/C. The field points away from the positive charge and towards the negative one.

16. 21-6 The Electric Field
Example 21-7: E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?
Figure Example 21–7. In (b), we don’t know the relative lengths of E1 and E2 until we do the calculation.
Solution: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x 108 N/C.
b. The acceleration is the force (charge times field) divided by the mass, and will be opposite to the direction of the field (due to the negative charge of the electron). Substitution gives a = 1.1 x 1020 m/s2.

17. 21-6 The Electric Field Example 21-8: above two point charges.
Calculate the total electric field (a) at point A and (b) at point B in the figure due to both charges, Q1 and Q2.
Solution: The geometry is shown in the figure. For each point, the process is: calculate the magnitude of the electric field due to each charge; calculate the x and y components of each field; add the components; recombine to give the total field.
a. E = 4.5 x 106 N/C, 76° above the x axis.
b. E = 3.6 x 106 N/C, along the x axis.

18. 21-6 The Electric Field
Problem solving in electrostatics: electric forces and electric fields
Draw a diagram; show all charges, with signs, and electric fields and forces with directions.
Calculate forces using Coulomb’s law.
Add forces vectorially to get result.
Check your answer!

19. ConcepTest 21.9b Superposition II4 3 2 1
-2 C
What is the electric field at the center of the square?
5) E = 0
Answer: 5

20. ConcepTest 21.9b Superposition II4 3 2 1
-2 C
What is the electric field at the center of the square?
5) E = 0
The four E field vectors all point outward from the center of the square toward their respective charges. Because they are all equal, the net E field is zero at the center!!
Follow-up: What if the upper two charges were +2 C? What if the right-hand charges were +2 C?

21. 21-7 Electric Field Calculations for Continuous Charge Distributions
A continuous distribution of charge may be treated as a succession of infinitesimal (point) charges. The total field is then the integral of the infinitesimal fields due to each bit of charge:
Remember that the electric field is a vector; you will need a separate integral for each component.

22. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m).
Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)

23. 21-7 Electric Field Calculations for Continuous Charge Distributions
Conceptual Example 21-10: Charge at the center of a ring.
Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces.
Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its equilibrium position, it will accelerate away from the ring.

24. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-11: Long line of charge.
Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m).
Solution: By symmetry, there will be no component of the field parallel to the wire. Write dE = (λ/4πε0) (cos θ dy)/x2 + y2). This can be transformed into an integral over θ: dE = (λ/4πε0x) cos θ dθ. Integrate θ from –π/2 to + π/2. Answer: E = (1/2πε0) λ/x.

25. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.
Solution: The disk is a set of concentric rings, and we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.

26. 21-7 Electric Field Calculations for Continuous Charge Distributions
In the previous example, if we are very close to the disk (that is, if z << R), the electric field is:
This is the field due to an infinite plane of charge.

27. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-13: Two parallel plates.
Determine the electric field between two large parallel plates or sheets, which are very thin and are separated by a distance d which is small compared to their height and width. One plate carries a uniform surface charge density σ and the other carries a uniform surface charge density -σ as shown (the plates extend upward and downward beyond the part shown).
Solution: The field due to each plate is σ/2ε0. Between the plates the fields add, giving E = σ/ε0; outside the plates the fields cancel.

28. 21-8 Field Lines
The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.
Figure Electric field lines (a) near a single positive point charge, (b) near a single negative point charge.

29. 21-8 Field Lines
The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge.
The electric field is stronger where the field lines are closer together.

30. 21-8 Field Lines
Electric dipole: two equal charges, opposite in sign:

31. ConcepTest 21.12b Electric Field Lines II1) 2)
3) both the same
Which of the charges has the greater magnitude?
Answer: 2

32. ConcepTest 21.12b Electric Field Lines II1) 2)
3) both the same
Which of the charges has the greater magnitude?
The field lines are denser around the red charge, so the red one has the greater magnitude.
Follow-up: What is the red/green ratio of magnitudes for the two charges?

33. 21-8 Field Lines
The electric field between two closely spaced, oppositely charged parallel plates is constant.
Field lines Demo

34. 21-8 Field Lines Summary of field lines:
Field lines indicate the direction of the field; the field is tangent to the line.
The magnitude of the field is proportional to the density of the lines.
Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

35. 21-9 Electric Fields and Conductors
The static electric field inside a conductor is zero – if it were not, the charges would move.
The net charge on a conductor resides on its outer surface.
Figure A charge inside a neutral spherical metal shell induces charge on its surfaces. The electric field exists even beyond the shell, but not within the conductor itself.

36. 21-9 Electric Fields and Conductors
The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.
Figure If the electric field at the surface of a conductor had a component parallel to the surface E||, the latter would accelerate electrons into motion. In the static case, E|| must be zero, and the electric field must be perpendicular to the conductor’s surface: E = E┴.

37. 21-9 Electric Fields and Conductors
Conceptual Example 21-14: Shielding, and safety in a storm.
A neutral hollow metal box is placed between two parallel charged plates as shown. What is the field like inside the box?
The field inside the box is zero. This is why it can be relatively safe to be inside an automobile during an electrical storm.

38. 21-10 Motion of a Charged Particle in an Electric Field
The force on an object of charge q in an electric field is given by:
= q
Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.

39. ConcepTest 21.6 Forces in 2D +2Q +4Q +Q 1 2 3 4 5
Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges?
Answer: 2

40. Follow-up: What would happen if the yellow charge were +3Q?
ConcepTest Forces in 2D
+2Q
+4Q +Q 1 2 3 4 5 d
Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges?
The charge +2Q repels +Q toward the right. The charge +4Q repels +Q upward, but with a stronger force. Therefore, the net force is up and to the right, but mostly up.
+2Q
Follow-up: What would happen if the yellow charge were +3Q?
+4Q

41. 21-10 Motion of a Charged Particle in an Electric Field
Example 21-15: Electron accelerated by electric field.
An electron (mass m = 9.11 x kg) is accelerated in the uniform field (E = 2.0 x 104 N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates.
Solution. a. The acceleration of the electron is qE/m = 3.5 x 1015 m/s2. In 1.5 cm it therefore accelerates from a speed of zero to v = 1.0 x 107 m/s.
b. The electric force on the electron is qE = 3.2 x N; the gravitational force is mg = 8.9 x N. Therefore the gravitational force can safely be ignored.

42. 21-10 Motion of a Charged Particle in an Electric Field
Example 21-16: Electron moving perpendicular to .
Suppose an electron traveling with speed v0 = 1.0 x 107 m/s enters a uniform electric field , which is at right angles to v0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.
Solution: The acceleration is in the vertical direction (perpendicular to the motion) and is equal to –eE/m. Then y = ½ ay2 and x = v0t; eliminating t gives the equation y = -(eE/2mv02)x2, which is a parabola.
Electrostatic Wind Demo

43. 21-11 Electric Dipoles
An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance . The dipole moment, p = Q , points from the negative to the positive charge.
Figure A dipole consists of equal but opposite charges, +Q and –Q, separated by a distance l. The dipole moment p + Ql and points from the negative to the positive charge.

44. 21-11 Electric Dipoles
An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:
Figure An electric dipole in a uniform electric field.

45. 21-11 Electric Dipoles
The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r3 dependence:
Figure Electric field due to an electric dipole.

46. 21-11 Electric Dipoles Example 21-17: Dipole in a field.
The dipole moment of a water molecule is 6.1 x C·m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 105 N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy when the torque is at its maximum? (c) In what position will the potential energy take on its greatest value? Why is this different than the position where the torque is maximum?
Solution. a. The maximum torque occurs then the field is perpendicular to the dipole moment; τ = pE = 1.2 x N·m.
b. The potential energy is zero.
c. The potential energy is maximum when the dipole moment is in the opposite direction to the field; this is the largest angle through which the dipole can rotate to reach equilibrium. The torque is maximum when the moment and the field are perpendicular. This situation is analogous to a physical pendulum – its gravitational potential energy is greatest when it is poised above the axis, even though the torque is zero there.

47. 21-12 Electric Forces in Molecular Biology; DNA
Molecular biology is the study of the structure and functioning of the living cell at the molecular level.
The DNA molecule is a double helix:
Figure 21-47a. Section of a DNA double helix.

48. 21-12 Electric Forces in Molecular Biology; DNA
The A-T and G-C nucleotide bases attract each other through electrostatic forces.
Figure 21-47b. “Close-up” view of the helix, showing how A and T attract each other and how G and C attract each other through electrostatic forces. The + and - signs represent net charges, usually a fraction of e, due to uneven sharing of electrons. The red dots indicate the electrostatic attraction (often called a “weak bond” or “hydrogen bond”—Section 40–3). Note that there are two weak bonds between A and T, and three between C and G.

49. 21-12 Electric Forces in Molecular Biology; DNA
Replication: DNA is in a “soup” of A, C, G, and T in the cell. During random collisions, A and T will be attracted to each other, as will G and C; other combinations will not.
Figure Replication of DNA.

50. 21-13 Photocopy Machines and Computer Printers Use Electrostatics
drum is charged positively
image is focused on drum
only black areas stay charged and therefore attract toner particles
image is transferred to paper and sealed by heat

51. 21-13 Photocopy Machines and Computer Printers Use Electrostatics
Figure Inside a photocopy machine: (1) the selenium drum is given (2) the lens focuses image on drum—only dark spots stay charged; (3) toner particles (negatively charged) are attracted to positive areas on drum; (4) the image is transferred to paper; (5) heat binds the image to the paper.

52. 21-13 Photocopy Machines and Computer Printers Use Electrostatics
Laser printer is similar, except a computer controls the laser intensity to form the image on the drum.
Figure Inside a laser printer: A movable mirror sweeps the laser beam in horizontal lines across the drum.

53. Summary of Chapter 21
Two kinds of electric charge – positive and negative.
Charge is conserved.
Charge on electron:
e = x C.
Conductors: electrons free to move.
Insulators: nonconductors.

54. Summary of Chapter 21 Charge is quantized in units of e.
Objects can be charged by conduction or induction.
Coulomb’s law:
Electric field is force per unit charge:

55. Summary of Chapter 21 Electric field of a point charge:
Electric field can be represented by electric field lines.
Static electric field inside conductor is zero; surface field is perpendicular to surface.