Chapter 18 Electric Forces and Electric Fields

Key Concepts:

electric charge•principle of conservation of charge•charge polarization, both permanent and induced•good electrical conductors vs. good electrical insulators•Coulomb's law for the force exerted by one charged particle on another

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the electric field concept; representation of an electric field using field lines or field vectors

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electric field of a point charge (formula and field pattern, both for a positive point charge and a negative point charge)

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field pattern for a constant (i.e., uniform) electric field•field pattern for an electric dipole•properties of electric field lines in the vicinity of conductors•

Chapter 18 LecturesJanuary-11-18 10:55 AM

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Q: Two metal rods have identical size, shape, and composition. Rod X has a net charge of +5 units, and

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composition. Rod X has a net charge of +5 units, and Rod Y has a net charge of −1 unit. The rods are originally separated, but then they are brought into contact with each other, while they are insulated from the rest of the world. After being in contact for some time, the rods are separated, and still insulated from the rest of the world. What is the final charge on each rod?

METAL RODS

Charge on X Charge on Y

BEFORE --------> +5 −1

AFTER (A.) +5 −1

AFTER (B.) +4 0

AFTER (C.) +3 +1

AFTER (D.) +2 +2

AFTER (E.) −1 +5

Q: Two plastic rods have identical size, shape, and composition. Rod X has a net charge of +5 units, and Rod Y has a net charge of −1 unit. The rods are originally separated, but then they are brought into contact with each other, while they are insulated from the rest of the world. After being in contact for some time, the rods are separated, and still insulated from the rest of the world. What is the final charge on each rod?

PLASTIC RODS

Charge on X Charge on Y

BEFORE --------> +5 −1

AFTER (A.) +5 −1

AFTER (B.) +4 0

AFTER (C.) +3 +1

AFTER (D.) +2 +2

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Coulomb's law

AFTER (D.) +2 +2

AFTER (E.) −1 +5

QUESTION: You have three metal rods with identical size, shape, and composition. Rod A has a charge of +6 units, and Rods B and C are neutral. Explain how to get a negative charge on one of the rods using only the three rods present.

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_________________________________________Q: In which case is the magnitude of the force on a positive charge located at the * position the greatest?

Q: In which case is the magnitude of the force on a positive charge located at the * position the least?

Discussion:

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Q: Determine the magnitude of the force that a proton exerts on an electron in a hydrogen atom. Assume that the distance between the particles is 5.29 × 10-11 m. The magnitude of the charge on each particle is 1.6 × 10-19 C.(A.) 5.1 × 1011 N(B.) 8.2 × 10-8 N(C.) 4.4 × 10-18 N

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(C.) 4.4 × 10-18 N(D.) 9.1 × 10-8 N(E.) [None of the above.]

Q: Two particles with positive charges QA and QB, where QA > QB, are held in place a distance r apart.

Which is greater, FAB or FBA?(A.) FAB > FBA (B.) FAB = FBA (C.) FAB < FBA__________________________________________

Q: In the previous question, how would the magnitude of the electric force on each particle change if the charge of one of the particles doubled and everything else remained the same?(A.) does not change(B.) increases by a factor of 2(C.) increases by a factor of 4(D.) decreases by a factor of 2(E.) [None of the above.]__________________________________________

Q: In the previous question, how would the magnitude of the electric force on each particle change if the distance between the particles doubled and everything else remained the same?(A.) does not change

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(A.) does not change(B.) increases by a factor of 2(C.) increases by a factor of 4(D.) decreases by a factor of 2(E.) [None of the above.]__________________________________________

Q: Consider the same situation as an earlier problem, but now include two additional particles so that the three "B" particles have the same charge and lie on the arc of a circle centred at the position of particle A, as in the following figure.

Now consider the following two explanations of the new situations:

Alice: "The net force on particle A due to the other three particles is now three times as great as before, because there is now three times as much charge producing the force, and the distances of the other three charges from charge A have not changed."

(A.) Alice is correct. (B.) Alice is incorrect.

Basil: "The force exerted on A by the top B particle cancels the force exerted on A by the bottom B particle , because they are symmetrically placed. Thus, the force exerted by the three B particles on A is the Ch18L Page 14

the force exerted by the three B particles on A is the same as in a previous situation, where there was only one B particle."

(A.) Basil is correct. (B.) Basil is incorrect. _____________________________________________

Q: Consider the following two situations. All of the particles are positively charged, and the magnitudes of all the charges on the particles labelled B are identical.

Alice: "The force exerted by the three B particles on Particle A in Situation 2 is three times as great as in Situation 1, because the charge of the three B particles is three times as great in Situation 2."

(A.) Alice is correct. (B.) Alice is incorrect.

Basil: "The force exerted on Particle A is the same in the two situations, because the forces from Particles B2 and B3 in Situation 2 are blocked by Particle B1."

(A.) Basil is correct. (B.) Basil is incorrect. ______________________________________________

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_____________________________________________ Q: The following diagram represents the electric field pattern for a particle that has a positive charge (i.e., a "point charge"). Is the magnitude of the electric field at position 1 greater than at point 2?

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The magnitude of the electric field is (A.) greater at position 1.(B.) greater at position 2.(C.) equal at the two positions. _____________________________________________

Principle of Superposition

The word "linear" is used to mean several different things in mathematics and physics. The classical theory of electricity and magnetism is a linear theory (at least in vacuum; in some materials it's a more complex story). This means that the fundamental differential equations of the theory are linear. This also means that a superposition principle is operative for electric fields.

That is, suppose you wish to calculate the net electric field due to several point charges. The principle of superposition states that it's possible to calculate the electric field due to each point charge separately, and then just add the resulting fields (vector sum) to determine the net electric field.

This is about as simple as it could possibly be, and makes our lives a lot easier. This is one reason why Einstein's theory of gravity is so hard to work with; it's a nonlinear theory, because its basic differential equations are nonlinear. Thus, if you use Einstein's theory of gravity to calculate the total gravitational field

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theory of gravity to calculate the total gravitational field due to two massive objects, you can't just find the field of each object separately and then form the vector sum to determine the net field. You have to treat each configuration of masses as a new problem that you have to solve "from scratch."

Newton's theory of gravity is a linear theory, so the superposition principle is valid for it. One way of looking at Newton's theory of gravity is that it's a linear approximation to Einstein's (nonlinear) theory of gravity. This perspective might appeal to all you calculus lovers out there.

The following example illustrates the superposition principle in action.

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_____________________________________________Q: The following diagram represents the electric field pattern for an electric dipole. Is the magnitude of the electric field at position 1 greater than at position 2?

The magnitude of the electric field is (A.) greater at position 1.(B.) greater at position 2.(C.) equal at the two positions._________________________________________Q: The following diagram represents the electric field pattern for a parallel-plate capacitor. Is the magnitude of the electric field at position 1 greater than at position 2?

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The magnitude of the electric field is (A.) greater at position 1.(B.) greater at position 2.(C.) equal at the two positions._______________________________________

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___________________________________________Q: Is the magnitude of the electric field at position 1 greater than at position 2?

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Example

An electric field of magnitude 100,000 N/C directed towards the right causes the 5.0 g ball in the figure to hang at a 20 degree angle. Determine the charge on the ball.

Q: Is the magnitude of the electric field at position 1 greater than at position 2?

The magnitude of the electric field is (A.) greater at position 1.(B.) greater at position 2.(C.) equal at the two positions._______________________________________

Additional Solved Problems

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Solution: Draw a free-body diagram:

The net force on the ball is zero, as the ball is permanently at rest. Write Newton's second law of motion for each coordinate direction in the free-body diagram:

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Problem

How many of the Earth's electrons would have to be transferred to the Sun so that the resulting electrostatic force of attraction between the Earth and Sun is equal to the gravitational force acting between Earth and Sun? What percentage of Earth's electrons does this represent?

Try this one on your own. You'll have to look up certain information; what specifically do you need?

We have two independent equations for two unknowns; eliminate T and solve for q:

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The mass of the Earth is about 1027 g, and Avogadro's number is about 1023, so this means that there are about 1050 atoms (very approximately), and so there are at least that many electrons in the Earth. You can see that the fraction of the Earth's electrons that would have to be transferred is extremely small as a percentage of the total number of electrons in the Earth.

You can do the calculation precisely; I'd be interested to see your solution.

Question: Plastic and glass rods that have been charged by rubbing with wool and silk, respectively, hang by threads. (a) An object repels the plastic rod. Can you predict what it will do to the glass rod? Explain. (b) Repeat part (a) for an object that attracts the plastic rod.

Selected problems and solutions:

My calculation yields a result of

Question: What is alike when we say "two like charges?" Do they look, feel, or smell alike?

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Question: A lightweight metal ball hangs by a thread. When a

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Question: A lightweight metal ball hangs by a thread. When a charged rod is brought near, the ball moves towards the rod, touches the rod, then quickly "flies away" from the rod. Explain.

Question: Metal sphere A has 4 units of negative charge, and metal sphere B has 2 units of positive charge. The two spheres are brought into contact. What is the final charge state of each sphere?

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Problem: A plastic rod has been charged to 20 nC by rubbing. (a) Have electrons been added or protons been removed? Explain. (b) How many electrons have been added or protons removed?

What is a coulomb?

Problem: A plastic rod that has been charged to 15.0 nC touches a metal sphere. Afterward, the charge on the rod

is 10.0 nC. (a) What kind of charged particle was transferred between the rod and the sphere, and in which direction? (b) How many charged particles were transferred?

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Problem: Two identical metal spheres A and B are connected by a metal rod. Both are initially neutral. Then 1.0 × 1012 electrons are added to sphere A, then the connecting rod is removed. Afterward, what is the charge on each sphere?

Problem: A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged rod is brought near (but not touching) the far end of A. While the charged rod is still close, A and B are separated. The charged rod is then withdrawn. Is the sphere then positively charged, negatively charged, or neutral? Explain.

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Question: Rank in order, from largest to smallest, noting any ties, the electric field strengths E1 to E4 at points 1 to 4 in the figure.

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Problem: Two small plastic spheres each have a mass of 2.0 g

and a charge of 50.0 nC. They are placed 2.0 cm apart. (a) What is the magnitude of the electric force between the two spheres? (b) By what factor is the electric force on a sphere larger than its weight?

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Problem: Determine the magnitude and direction of the force on charge A in the figure.

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Notice how we used Coulomb's law to determine the magnitudes of the forces, and then inserted the positive and negative signs for the directions of the forces "by hand" at the end._________________________________________________

Problem: What magnitude charge creates a 1.0 N/C electric field at a point 1.0 m away?

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Problem: A 30 nC charged particle and a 50 nC charged particle are near each other. There are no other charges nearby. The electric force on the 30 nC particle is 0.035 N. The 50 nC particle is then moved very far away. Afterward, what is the magnitude of the electric field at its original position?

Problem: A +10 nC charge is located at the origin. (a) What are the strengths of the electric field vectors at the positions (x, y) =

(5.0 cm, 0.0 cm), (5.0 cm, 5.0 cm), and (5.0 cm, 5.0 cm)? (b) Draw a diagram showing the electric field vectors at these points.

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points.

Solution:

Problem: What is the strength of an electric field that will balance the weight of a 1.0 g plastic sphere that has

been charged to 3.0 nC?

Solution:

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Problem: A parallel-plate capacitor is formed from two 4.0 cm × 4.0 cm electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0 × 106

N/C. What is the charge in nC on each electrode?

Solution:

Problem: A 2.0-mm-diameter copper ball is charged to +50 nC. What fraction of its electrons have been removed? Copper has density 8900 kg/m3, atomic mass 63.5 g/mole, and atomic number 29.

Solution:

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Problem: Two protons are 2.0 fm apart. Determine the magnitude of the (a) electric force and (b) gravitational force of one proton on the other. (c) Calculate the ratio of the electric force to the gravitational force and comment.

Solution:

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Problem: An electric dipole is formed from 1.0 nC point charges spaced 2.0 mm apart. The dipole is centred at the origin, oriented along the y-axis. Determine the electric field strengths at the points (a) (x, y) = (10 mm, 0 mm) and (b) (x, y) = (0 mm, 10 mm).

Solution:

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Problem: The figure shows four charges at the corners of a square of side L. Assume that q and Q are positive. (a) Draw a diagram showing the three forces on charge q due to the other charges. (b) Obtain an expression for the magnitude of the net force on q.

Solution:

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Problem: Two 2.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ±10 nC. (a) Determine the electric field strength between the disks. (b) A proton is shot from the negative disk towards the positive disk. What launch speed must the proton have to just barely reach the positive disk?

Solution:

.

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Problem: A small bead with a positive charge q is free to slide on a horizontal wire of length 4.0 cm. At the left end of the wire is a fixed charge q, and at the right end is a fixed charge 4q. How far from the left end of the wire does the bead come to rest?

Solution:

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Problem: An electric field E = (200,000 N/C, right) causes the 2.0 g ball in the figure to hang at an angle. Calculate the angle θ.

Solution:

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Problem: An electron is released from rest at the negative plate of a parallel plate capacitor. The magnitude of the charge per unit area on each plate is 1.8 × 10-7 C/m2, and the plates are separated by a distance of 1.5 × 10-2 m. How fast is the electron moving just before it reaches the positive plate?

Solution: Assume that the electric field inside the capacitor is uniform (standard assumption for a parallel-plate capacitor). The force exerted by the capacitor's electric field on the electric field is therefore constant, and the electron moves in a straight line across the capacitor.

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The magnitude of the capacitor's uniform electric field is:

By Newton's second law, the constant acceleration of the electron is:

Using the following kinematics equation, we can determine the speed of the electron just before it hits the positive plate.

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Problem: Two particles are in a uniform electric field whose magnitude is 2500 N/C. The mass and charge of particle 1 are m1 = 1.4 × 10-5 kg and q1 = -7.0 × 10-6 C, whereas the corresponding values for particle 2 are m2 = 2.6 × 10-5 kg and q2 = +18 × 10-6 C. Initially the particles are at rest. The particles are both located on the same electric field line but are separated from each other by a distance d. When released, they accelerate, but always remain at this same distance from each other. Find d.

Solution: Draw a free-body diagram for each particle. Then use the condition that the acceleration of each particle is the same, in conjunction with Newton's second law. You will be left with one equation with only one unknown, d, which can be solved for d.

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Problem: The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 7.00 × 106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

(That's a very big capacitor!)

Do you understand why the values for the electric charges of the two particles were inserted into the equation as positive numbers, even though one of the charges was negative? Remember that when we draw free-body diagrams, we put the directions of the forces in "by hand" and then input the values into the resulting equations as magnitudes only.

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Solution: Doesn't this problem remind you of projectile motion? Yes! It's the same idea. The path of the electron, which is a curve, might be intimidating at first, but when you realize that you can treat the components of the motion of the electron separately, it is less intimidating.

Set up your coordinate system so that the positive x-axis is to the right and the positive y-axis is upward. Then the force in the x-direction is zero, which makes that component easy. The force in the y-direction is constant, so that component is also pretty easy.

Here's my strategy for solving the problem: Use the dimensions given and kinematics to determine the acceleration of the electron. Then use Newton's second law to determine the force acting on the electron, which can then be connected to the electric field.

(See important note at end of solution.)

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The magnitude of the electric field is 2100 N/C. The direction of the electric field is downward (i.e., in the negative y-direction).________________________________________________________

Important note about calculators

You will have noticed that I have the habit of not using my calculator until the very last calculation, unless it's absolutely necessary. For instance, in the first calculation in the solution to this problem, I did the following:

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Notice that I did not use my calculator at this point. Some students did use their calculator at this point when solving this problem on a test; their result was

0.000000002

DON’T DO THIS! The safest strategy is to leave intermediate results in exact form. However, if you must calculate intermediate results, remember to use more decimal places than you will ultimately need.

The problem here is that the calculator has TRUNCATED; that is, it only shows the first 10 digits in a calculation, WITHOUT ANY ROUNDING. It just doesn't display any subsequent digits; this is the meaning of the word "truncate."

If you multiply the result of your calculation by 1,000,000,000, you'll obtain

2.857142857

Ah, there are more digits after the initial one! The calculator only displayed the first digit, and if you only recorded this first digit you have made quite a substantial rounding error (30% in fact).

Be aware of this problem; a good way to avoid making such rounding errors is to place your calculator in a different mode, such as scientific mode (look for the FSE key, the third from the left in the top row of the calculator that many of you use, the SHARP DAL EL 510R or EL 510RN; this is a toggle key, and by pressing it you can toggle between the three modes). If you place your calculator in scientific notation mode, you will not make such rounding errors.

Be careful! Calculators have no intelligence, but you do! As you become more and more aware of the capabilities and limitations of your tools, you become a more and more intelligent wielder of them.

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them.

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