Chapter 21 Electric Charge and Electric Field

Copyright © 2009 Pearson Education, Inc.

Chapter 21Electric Charge and

Electric Field

Two balls with charges Two balls with charges ++QQ and and +4+4QQ are separated by are separated by 33RR. Where . Where

should you place another charged ball should you place another charged ball QQ00 on the line between on the line between

the two charges such that the net force on the two charges such that the net force on QQ00 will be zero? will be zero?

3R

+Q +4Q

R 2R

1 2 3 4 5

ConcepTest 21.4bConcepTest 21.4b Electric Force IIElectric Force II

Two balls with charges Two balls with charges ++QQ and and +4+4QQ are separated by are separated by 33RR. Where . Where

should you place another charged ball should you place another charged ball QQ00 on the line between on the line between

the two charges such that the net force on the two charges such that the net force on QQ00 will be zero? will be zero?

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.

3R

+Q +4Q

R 2R

1 2 3 4 5

ConcepTest 21.4bConcepTest 21.4b Electric Force IIElectric Force II

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

p e

ConcepTest 21.5cConcepTest 21.5c Proton and Electron IIIProton and Electron III

By Newton’s 3rd law, the electron and proton feel the same forcesame force. But, since F = ma, and since the proton’s mass is much greaterproton’s mass is much greater, the proton’s acceleration will be much smallerproton’s acceleration will be much smaller!

Thus, they will meet closer to the proton’s closer to the proton’s original positionoriginal position.

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

p e

ConcepTest 21.5cConcepTest 21.5c Proton and Electron IIIProton and Electron III

Follow-up:Follow-up: Which particle will be moving faster when they meet? Which particle will be moving faster when they meet?


For a point charge:

21-6 The Electric Field



An electric field surrounds every charge.


Force on a point charge in an 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.


21-6 The Electric FieldExample 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 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?

43

2 1

-2 C

-2 C -2 C

-2 C

ConcepTest 21.9bConcepTest 21.9b Superposition IISuperposition II

What is the electric field at What is the electric field at

the center of the square?the center of the square?

1) E = 02) E = 13) E = 24) E = 35) E = 4

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 net EE field is zero at the center field is zero at the center!!

43

2 1

-2 C

-2 C -2 C

-2 C

ConcepTest 21.9bConcepTest 21.9b Superposition IISuperposition II

What is the electric field at What is the electric field at

the center of the square?the center of the square?

5) E = 0

Follow-up:Follow-up: What if the upper two charges were +2 C? What if the upper two charges were +2 C? What if the right-hand charges were +2 C? What if the right-hand charges were +2 C?


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.



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).



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).

SYMMETRY!!



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.



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).



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).

SYMMETRY!!



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.



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.

SYMMETRY!!



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.



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).


The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

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.

21-8 Field Lines

Schematic only – fields in 3-d, not 2-d


Electric dipole: two equal charges, opposite in sign:

21-8 Field Lines

ConcepTest 21.12bConcepTest 21.12b Electric Field Lines IIElectric Field Lines II

Which of the charges has

the greater magnitude?

1)

2)

3) both the same

ConcepTest 21.12bConcepTest 21.12b Electric Field Lines IIElectric Field Lines II

Which of the charges has

the greater magnitude?

1)

2)

3) both the same

The field lines are denser around denser around

the red chargethe red charge, so the red one red one

has the greater magnitudehas the greater magnitude.

Follow-up:Follow-up: What is the red/green ratio What is the red/green ratio of magnitudes for the two charges?of magnitudes for the two charges?


The electric field between two closely spaced, oppositely charged parallel plates is constant.

21-8 Field Lines


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.

21-8 Field Lines


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.

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.




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?


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.

E��

E��

F��

Which of the arrows best

represents the direction

of the net force on charge

+Q due to the other two

charges?

+2Q

+4Q

+Q

1 23

4

5d

d

ConcepTest 21.6ConcepTest 21.6 Forces in 2DForces in 2D

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 net force is up and to

the right, but mostly upthe right, but mostly up.

+2Q

+4Q

+Q

1 23

4

5d

d

+2Q

+4Q

ConcepTest 21.6ConcepTest 21.6 Forces in 2DForces in 2D

Which of the arrows best

represents the direction

of the net force on charge

+Q due to the other two

charges?

Follow-up:Follow-up: What would happen if What would happen if the yellow charge were +3the yellow charge were +3QQ??



Example 21-15: Electron accelerated by electric field.

An electron (mass m = 9.11 x 10-31 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.

E��



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.

E

E


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.

��

p



An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:



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:


21-11 Electric DipolesExample 21-17: Dipole in a field.

The dipole moment of a water molecule is 6.1 x 10-30 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?


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:

21-12 Electric Forces in Molecular Biology; DNA


The A-T and G-C nucleotide bases attract each other through electrostatic forces.



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.



• Two kinds of electric charge – positive and negative.

• Charge is conserved.

• Charge on electron:

e = 1.602 x 10-19 C.

• Conductors: electrons free to move.

• Insulators: nonconductors.

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:



• 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.



Chapter 22Gauss’s Law


• Electric Flux

• Gauss’s Law

• Applications of Gauss’s Law

• Experimental Basis of Gauss’s and Coulomb’s Laws

Units of Chapter 22


Electric flux:

Electric flux through an area is proportional to the total number of field lines crossing the area.

22-1 Electric Flux


22-1 Electric Flux

Example 22-1: Electric flux.

Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.


Flux through a closed surface:

22-1 Electric Flux


Think in terms of flux lines:


The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law:

This can be used to find the electric field in situations with a high degree of symmetry.

22-2 Gauss’s Law


22-2 Gauss’s Law

For a point charge,

Therefore,

Solving for E gives the result we expect from Coulomb’s law:

Spherical


22-2 Gauss’s LawUsing Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives:

Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.


22-2 Gauss’s Law

Finally, if a gaussian surface encloses several point charges, the superposition principle shows that:

Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.


22-2 Gauss’s Law

Conceptual Example 22-2: Flux from Gauss’s law.

Consider the two gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2?


22-3 Applications of Gauss’s LawExample 22-3: Spherical conductor.

A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Spherical


22-3 Applications of Gauss’s Law

Example 22-4: Solid sphere of charge.

An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).

Spherical/radial



Example 22-5: Nonuniformly charged solid sphere.

Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.



Example 22-6: Long uniform line of charge.

A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends.

Linear,Cylindrical



Example 22-7: Infinite plane of charge.

Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane.

Planar & cylindrical



Example 22-8: Electric field near any conducting surface.

Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by

E = σ/ε0

where σ is the surface charge density on the conductor’s surface at that point.



The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways:

1. The field inside the conductor is zero, so the flux is all through one end of the cylinder.

2. The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.



Conceptual Example 22-9: Conductor with charge inside a cavity.

Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor?



Procedure for Gauss’s law problems:

1. Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field).

2. Draw the surface.

3. Use the symmetry to find the direction of E.

4. Evaluate the flux by integrating.

5. Calculate the enclosed charge.

6. Solve for the field.

��E


22-4 Experimental Basis of Gauss’s and Coulomb’s Laws

In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero.


• Electric flux:

• Gauss’s law:

• Gauss’s law can be used to calculate the field in situations with a high degree of symmetry.

• Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law.



Homework Asignment # 2

Chapter 21 – 40, 58, 62Chapter 22 – 10, 22, 34

Homework Asignment # 3

Chapter 23 – 4, 14, 30, 36, 52, 62

Documents

Chapter 21 Electric Charge and Electric Field