Chapter 21 Electric Charge and Electric Field

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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.

Units of Chapter 21 Static Electricity; Electric Charge and Its Conservation Electric Charge in the Atom Insulators and Conductors Induced Charge; the Electroscope Coulomb’s Law The Electric Field Electric Field Calculations for Continuous Charge Distributions

Units of Chapter 21 Field Lines Electric Fields and Conductors
Motion of a Charged Particle in an Electric Field Electric Dipoles Electric Forces in Molecular Biology: DNA Photocopy Machines and Computer Printers Use Electrostatics

21-1 Static Electricity; Electric Charge and Its Conservation
Objects can be charged by rubbing Figure (a) Rub a plastic ruler and (b) bring it close to some tiny pieces of paper.

21-1 Static Electricity; Electric Charge and Its Conservation
Charge comes in two types, positive and negative; like charges repel and opposite charges attract. Figure Like charges repel one another; unlike charges attract. (Note color coding: positive and negative charged objects are colored rose-pink and blue-green, respectively, in this book.)

ConcepTest 21.1a Electric Charge I
1) one is positive, the other is negative 2) both are positive 3) both are negative 4) 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? Answer: 4

ConcepTest 21.1a Electric Charge I
1) one is positive, the other is negative 2) both are positive 3) both are negative 4) 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? 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. Follow-up: What does the picture look like if the two balls are oppositely charged? What about if both balls are neutral?

ConcepTest 21.1b Electric Charge II
1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? Answer: 3

ConcepTest 21.1b Electric Charge II
1) have opposite charges 2) have the same charge 3) all have the same charge 4) 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).

21-1 Static Electricity; Electric Charge and Its Conservation
Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction.

21-2 Electric Charge in the Atom
Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge) Figure Simple model of the atom.

21-2 Electric Charge in the Atom
Polar molecule: neutral overall, but charge not evenly distributed Figure Diagram of a water molecule. Because it has opposite charges on different ends, it is called a “polar” molecule.

21-3 Insulators and Conductors
Charge flows freely Metals Insulator: Almost no charge flows Most other materials Some materials are semiconductors. Figure (a) A charged metal sphere and a neutral metal sphere. (b) The two spheres connected by a conductor (a metal nail), which conducts charge from one sphere to the other. (c) The two spheres connected by an insulator (wood); almost no charge is conducted.

21-4 Induced Charge; the Electroscope
Metal objects can be charged by conduction: Figure A neutral metal rod in (a) will acquire a positive charge if placed in contact (b) with a positively charged metal object. (Electrons move as shown by the orange arrow.) This is called charging by conduction.

21-4 Induced Charge; the Electroscope
They can also be charged by induction, either while connected to ground or not: Figure Charging by induction. Figure Inducing a charge on an object connected to ground.

21-4 Induced Charge; the Electroscope
Nonconductors won’t become charged by conduction or induction, but will experience charge separation: Figure A charged object brought near an insulator causes a charge separation within the insulator’s molecules.

ConcepTest 21.2a Conductors I
1) positive 2) negative 3) neutral 4) positive or neutral 5) 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: Answer: 5

ConcepTest 21.2a Conductors I
1) positive 2) negative 3) neutral 4) positive or neutral 5) 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: 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. Remember the ball is a conductor! Follow-up: What happens if the metal ball is replaced by a plastic ball?

21-4 Induced Charge; the Electroscope
The electroscope can be used for detecting charge. Figure Electroscope.

21-4 Induced Charge; the Electroscope
The electroscope can be charged either by conduction or by induction. Figure Electroscope charged (a) by induction, (b) by conduction.

21-4 Induced Charge; the Electroscope
The charged electroscope can then be used to determine the sign of an unknown charge. Figure A previously charged electroscope can be used to determine the sign of a charged object.

ConcepTest 21.2b Conductors II
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? 1) 0 0 2) + – 3) – + 4) + + 5) – – ? Answer: 3

ConcepTest 21.2b Conductors II
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? 1) 0 0 2) + – 3) – + 4) + + 5) – – ? 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. Follow-up: What will happen when the conductors are reconnected with a wire?

21-5 Coulomb’s Law 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. Figure Coulomb’s law, Eq. 21–1, gives the force between two point charges, Q1 and Q2, a distance r apart.

21-5 Coulomb’s Law Coulomb’s law:
This equation gives the magnitude of the force between two charges.

21-5 Coulomb’s Law The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same. Figure The direction of the static electric force one point charge exerts on another is always along the line joining the two charges, and depends on whether the charges have the same sign as in (a) and (b), or opposite signs (c).

ConcepTest 21.3a Coulomb’s Law I
What is the magnitude of the force F2? Q F1 = 3 N F2 = ? Answer: 4

ConcepTest 21.3a Coulomb’s Law I
What is the magnitude of the force F2? Q F1 = 3 N F2 = ? The force F2 must have the same magnitude as F1. This is due to the fact that the form of Coulomb’s law is totally symmetric with respect to the two charges involved. The force of one on the other of a pair is the same as the reverse. Note that this sounds suspiciously like Newton’s 3rd law!!

ConcepTest 21.3b Coulomb’s Law II
Q F1 = 3 N F2 = ? If we increase one charge to 4Q, what is the magnitude of F1? 4Q Q F1 = ? F2 = ? Answer: 3

ConcepTest 21.3b Coulomb’s Law II
Q F1 = 3 N F2 = ? If we increase one charge to 4Q, what is the magnitude of F1? 4Q Q F1 = ? F2 = ? Originally we had: F1 = k(Q)(Q)/r2 = 3 N Now we have: F1 = k(4Q)(Q)/r2 which is 4 times bigger than before. Follow-up: Now what is the magnitude of F2?

ConcepTest 21.3c Coulomb’s Law III
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? 1) 9F 2) 3F 3) F 4) 1/3F 5) 1/9F Q F d ? 3d Answer: 5

ConcepTest 21.3c Coulomb’s Law III
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? 1) 9F 2) 3F 3) F 4) 1/3F 5) 1/9F Q F d Originally we had: Fbefore = k(Q)(Q)/d2 = F Now we have: Fafter = k(Q)(Q)/(3d)2 = 1/9F Q ? 3d Follow-up: What is the force if the original distance is halved?

21-5 Coulomb’s Law Unit of charge: coulomb, C.
The proportionality constant in Coulomb’s law is then: k = 8.99 x 109 N·m2/C2. Charges produced by rubbing are typically around a microcoulomb: 1 μC = 10-6 C.

21-5 Coulomb’s Law Charge on the electron: e = 1.602 x 10-19 C.
Electric charge is quantized in units of the electron charge.

21-5 Coulomb’s Law The proportionality constant k can also be written in terms of ε0, the permittivity of free space:

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.

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).

ConcepTest 21.4a Electric Force I
Two balls with charges +Q and +4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 1) yes, but only if Q0 is positive 2) yes, but only if Q0 is negative 3) yes, independent of the sign (or value) of Q0 4) no, the net force can never be zero 3R +Q +4Q Answer: 3

ConcepTest 21.4a Electric Force I
Two balls with charges +Q and +4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 1) yes, but only if Q0 is positive 2) yes, but only if Q0 is negative 3) yes, independent of the sign (or value) of Q0 4) no, the net force can never be zero A positive charge would be repelled by both charges, so a point where these two repulsive forces cancel can be found. A negative charge would be attracted by both, and the same argument holds. 3R +Q +4Q Follow-up: What happens if both charges are +Q? Where would the F = 0 point be in this case?

ConcepTest 21.4c Electric Force III
Two balls with charges +Q and –4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q0 anywhere on the line such that the net force on Q0 will be zero? 1) yes, but only if Q0 is positive 2) yes, but only if Q0 is negative 3) yes, independent of the sign (or value) of Q0 4) no, the net force can never be zero 3R +Q – 4Q Answer: 3

ConcepTest 21.4c Electric Force III
Two balls with charges +Q and –4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q0 anywhere on the line such that the net force on Q0 will be zero? 1) yes, but only if Q0 is positive 2) yes, but only if Q0 is negative 3) yes, independent of the sign (or value) of Q0 4) no, the net force can never be zero A charge (positive or negative) can be placed to the left of the +Q charge, such that the repulsive force from the +Q charge cancels the attractive force from –4Q. 3R +Q – 4Q Follow-up: What happens if one charge is +Q and the other is –Q?

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.

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.

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

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

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.

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

21-6 The Electric Field For a point charge:

ConcepTest 21.7 Electric Field
You are sitting a certain distance from a point charge, and you measure an electric field of E0. If the charge is doubled and your distance from the charge is also doubled, what is the electric field strength now? 1) 4E0 2) 2E0 3) E0 4) 1/2E0 5) 1/4E0 Answer: 4

ConcepTest 21.7 Electric Field
You are sitting a certain distance from a point charge, and you measure an electric field of E0. If the charge is doubled and your distance from the charge is also doubled, what is the electric field strength now? 1) 4E0 2) 2E0 3) E0 4) 1/2E0 5) 1/4E0 Remember that the electric field is: E = kQ/r2. Doubling the charge puts a factor of 2 in the numerator, but doubling the distance puts a factor of 4 in the denominator, because it is distance squared!! Overall, that gives us a factor of 1/2. Follow-up: If your distance is doubled, what must you do to the charge to maintain the same E field at your new position?

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.

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.

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.

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.

ConcepTest 21.9a Superposition I
4 3 2 1 -2 C What is the electric field at the center of the square? 5) E = 0 Answer: 3

ConcepTest 21.9a Superposition I
4 3 2 1 -2 C What is the electric field at the center of the square? 5) E = 0 For the upper charge, the E field vector at the center of the square points toward that charge. For the lower charge, the same thing is true. Then the vector sum of these two E field vectors points to the left. Follow-up: What if the lower charge were +2 C? What if both charges were +2 C?

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!

Summary of Chapter 21 Sec. 1-6
Two kinds of electric charge – positive and negative. Charge is conserved. Charge on electron: e = x C. Conductors: electrons free to move. Insulators: nonconductors.

Summary of Chapter 21 Sec. 1-6
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:

Summary of Chapter 21 Sec. 1-6
Electric field of a point charge: