Copyright © 2009 Pearson Education, Inc. Chapter 21 Electric Charge and Electric Field

Copyright © 2009 Pearson Education, Inc. Objects can be charged by rubbing Static Electricity; Electric Charge and Its Conservation

Copyright © 2009 Pearson Education, Inc. Charge comes in two types, positive and negative; like charges repel and opposite charges attract Static Electricity; Electric Charge and Its Conservation

Copyright © 2009 Pearson Education, Inc. Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction Static Electricity; Electric Charge and Its Conservation

Copyright © 2009 Pearson Education, Inc. Atom: Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge) Electric Charge in the Atom

Copyright © 2009 Pearson Education, Inc. Polar molecule: neutral overall, but charge not evenly distributed Electric Charge in the Atom

Copyright © 2009 Pearson Education, Inc. Conductor: Charge flows freely Metals Insulator: Almost no charge flows Most other materials Some materials are semiconductors Insulators and Conductors

Copyright © 2009 Pearson Education, Inc. Metal objects can be charged by conduction: 21-4 Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. They can also be charged by induction, either while connected to ground or not: 21-4 Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. Nonconductors won’t become charged by conduction or induction, but will experience charge separation: 21-4 Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. The electroscope can be used for detecting charge. The more charge there is, the more the leaves are deflected Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. The electroscope can be charged either by conduction or by induction Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. The charged electroscope can then be used to determine the sign of an unknown charge Induced Charge; the Electroscope

Copyright © 2009 Pearson Education, Inc. Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them Coulomb’s Law

Copyright © 2009 Pearson Education, Inc. Coulomb’s law: This equation gives the magnitude of the force between two charges Coulomb’s Law

Copyright © 2009 Pearson Education, Inc. The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same Coulomb’s Law

Copyright © 2009 Pearson Education, Inc. Unit of charge: coulomb, C. The proportionality constant in Coulomb’s law is then: k = 8.99 x 10 9 N·m 2 /C 2. Charges produced by rubbing are typically around a microcoulomb: 1 μC = C Coulomb’s Law

Copyright © 2009 Pearson Education, Inc. Charge on the electron: e = x C. Electric charge is quantized in units of the electron charge Coulomb’s Law

Copyright © 2009 Pearson Education, Inc. The proportionality constant k can also be written in terms of ε 0, the permittivity of free space: 21-5 Coulomb’s Law

Copyright © 2009 Pearson Education, Inc Coulomb’s Law Conceptual Example 21-1: Which charge exerts the greater force? Two positive point charges, Q 1 = 50 μC and Q 2 = 1 μC, are separated by a distance L. Which is larger in magnitude, the force that Q 1 exerts on Q 2 or the force that Q 2 exerts on Q 1 ?

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc Coulomb’s Law Example 21-3: Electric force using vector components. Calculate the net electrostatic force on charge Q 3 shown in the figure due to the charges Q 1 and Q 2.

Copyright © 2009 Pearson Education, Inc. The electric field is defined as the force on a small charge, divided by the magnitude of the charge: 21-6 The Electric Field

Copyright © 2009 Pearson Education, Inc The Electric Field An electric field surrounds every charge.

Copyright © 2009 Pearson Education, Inc. For a point charge: 21-6 The Electric Field

Copyright © 2009 Pearson Education, Inc. Force on a point charge in an electric field: 21-6 The Electric Field

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc 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 C.

Copyright © 2009 Pearson Education, Inc 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)?

Copyright © 2009 Pearson Education, Inc 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, Q 1 and Q 2.

Copyright © 2009 Pearson Education, Inc. Problem solving in electrostatics: electric forces and electric fields 1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions. 2. Calculate forces using Coulomb’s law. 3. Add forces vectorially to get result. 4. Check your answer! 21-6 The Electric Field

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc 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).

Copyright © 2009 Pearson Education, Inc 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).

Copyright © 2009 Pearson Education, Inc 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/m 2 ) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc 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).

Copyright © 2009 Pearson Education, Inc. The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge Field Lines

Copyright © 2009 Pearson Education, Inc. 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 Field Lines

Copyright © 2009 Pearson Education, Inc. Electric dipole: two equal charges, opposite in sign: 21-8 Field Lines

Copyright © 2009 Pearson Education, Inc. The electric field between two closely spaced, oppositely charged parallel plates is constant Field Lines

Copyright © 2009 Pearson Education, Inc. Summary of field lines: 1.Field lines indicate the direction of the field; the field is tangent to the line. 2.The magnitude of the field is proportional to the density of the lines. 3.Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge Field Lines

Copyright © 2009 Pearson Education, Inc. 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 Electric Fields and Conductors

Copyright © 2009 Pearson Education, Inc. The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move Electric Fields and Conductors

Copyright © 2009 Pearson Education, Inc 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?

Copyright © 2009 Pearson Education, Inc Motion of a Charged Particle in an Electric Field The force on an object of charge q in an electric field E is given by: F = qE Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.

Copyright © 2009 Pearson Education, Inc 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 10 4 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.

Copyright © 2009 Pearson Education, Inc Motion of a Charged Particle in an Electric Field Example 21-16: Electron moving perpendicular to. Suppose an electron traveling with speed v 0 = 1.0 x 10 7 m/s enters a uniform electric field, which is at right angles to v 0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.

Copyright © 2009 Pearson Education, Inc 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.

Copyright © 2009 Pearson Education, Inc Electric Dipoles An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:

Copyright © 2009 Pearson Education, Inc 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/r 3 dependence:

Copyright © 2009 Pearson Education, Inc 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 10 5 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?

Copyright © 2009 Pearson Education, Inc. 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: Electric Forces in Molecular Biology; DNA

Copyright © 2009 Pearson Education, Inc. The A-T and G-C nucleotide bases attract each other through electrostatic forces Electric Forces in Molecular Biology; DNA

Copyright © 2009 Pearson Education, Inc. 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 Electric Forces in Molecular Biology; DNA

Copyright © 2009 Pearson Education, Inc. Photocopy machine: 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 Photocopy Machines and Computer Printers Use Electrostatics

Copyright © 2009 Pearson Education, Inc Photocopy Machines and Computer Printers Use Electrostatics

Copyright © 2009 Pearson Education, Inc. Laser printer is similar, except a computer controls the laser intensity to form the image on the drum Photocopy Machines and Computer Printers Use Electrostatics

Copyright © 2009 Pearson Education, Inc. Review Questions

ConcepTest 21.1aElectric Charge I ConcepTest 21.1a Electric Charge I A) one is positive, the other is negative B) both are positive C) both are negative D) both are positive or both are negative Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges?

ConcepTest 21.1aElectric Charge I ConcepTest 21.1a Electric Charge I same charge The fact that the balls repel each other can tell you only that they have the same charge, but you do not know the sign. So they can be either both positive or both negative. A) one is positive, the other is negative B) both are positive C) both are negative D) both are positive or both are negative Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges? Follow-up: What does the picture look like if the two balls are oppositely charged? What about if both balls are neutral?

A) have opposite charges B) have the same charge C) all have the same charge D) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? ConcepTest 21.1bElectric Charge II ConcepTest 21.1b Electric Charge II

A) have opposite charges B) have the same charge C) all have the same charge D) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? The GREEN and PINK balls must have the same charge, since they repel each other. The YELLOW ball also repels the GREEN, so it must also have the same charge as the GREEN (and the PINK). ConcepTest 21.1bElectric Charge II ConcepTest 21.1b Electric Charge II

ConcepTest 21.2aConductors I ConcepTest 21.2a Conductors I A) positive B) negative C) neutral D) positive or neutral E) negative or neutral A metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positive-charged rod held near the ball. The charge of the ball must be:

negative neutral induction Clearly, the ball will be attracted if its charge is negative. However, even if the ball is neutral, the charges in the ball can be separated by induction (polarization), leading to a net attraction. A) positive B) negative C) neutral D) positive or neutral E) negative or neutral A metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positive-charged rod held near the ball. The charge of the ball must be: Remember the ball is a conductor! ConcepTest 21.2aConductors I ConcepTest 21.2a Conductors I Follow-up: What happens if the metal ball is replaced by a plastic ball?

Two neutral conductors are connected by a wire and a charged rod is brought near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? ConcepTest 21.2bConductors II ConcepTest 21.2b Conductors II A)00 B)+– C)–+ D)++ E)– – 0 0 ? ?

positive charge will flow from the blue to the green ball due to polarization charges will remain on the separate conductors While the conductors are connected, positive charge will flow from the blue to the green ball due to polarization. Once disconnected, the charges will remain on the separate conductors even when the rod is removed. Two neutral conductors are connected by a wire and a charged rod is brought near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? ConcepTest 21.2bConductors II ConcepTest 21.2b Conductors II A)00 B)+– C)–+ D)++ E)– – 0 0 ? ? Follow-up: What will happen when the conductors are reconnected with a wire?

A) 9F B) 3F C) F D) 1/3F E) 1/9F The force between two charges separated by a distance d is F. If the charges are pulled apart to a distance 3d, what is the force on each charge? QF QFd Q ? Q ? 3d3d3d3d ConcepTest 21.3cCoulomb’s Law III ConcepTest 21.3c Coulomb’s Law III

Originally we had: F before = k(Q)(Q)/d 2 = F Now we have: F after = k(Q)(Q)/(3d) 2 = 1/9F A) 9F B) 3F C) F D) 1/3F E) 1/9F The force between two charges separated by a distance d is F. If the charges are pulled apart to a distance 3d, what is the force on each charge? QF QFd Q?Q? 3d3d3d3d ConcepTest 21.3cCoulomb’s Law III ConcepTest 21.3c Coulomb’s Law III Follow-up: What is the force if the original distance is halved?

Copyright © 2009 Pearson Education, Inc. Assignments Group problems: 6, 8, 16, 22 HW Problems: #’s 7, 9, 13, 15, 25, 27, 29, 31, 33, 37, 41, 45, 57, 63 Read back over Ch. 21 and start going through Ch. 22.