electric charge and electric field

Electric Charge and Electric Field

• 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

• 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

Charge comes in two types, positive and negative; like charges repel and opposite charges attract.

Electric Charge and Its

Conservation

Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction.

Electric Charge and Its Conservation

Atom:

Nucleus (small, massive, positive charge)

Electron cloud (large, very low density, negative charge)

Electric Charge in the Atom

Polar molecule: neutral overall, but charge not evenly distributed

Electric Charge in the Atom

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:

Induced Charge

Copyright © 2009 Pearson Education, Inc.

They can also be charged by induction, either while connected to ground or not:

Induced Charge

Copyright © 2009 Pearson Education, Inc.

Nonconductors won’t become charged by conduction or induction, but will experience charge separation:

Induced Charge

Copyright © 2009 Pearson Education, Inc.

The electroscope can be used for detecting charge.

the Electroscope

Electric Charge IElectric 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 Two charged balls are

repelling each other as repelling each other as

they hang from the ceiling. they hang from the ceiling.

What can you say about What can you say about

their charges?their charges?

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, From the picture,

what can you what can you

conclude about conclude about

the charges?the charges?

Electric Charge IIElectric Charge II

Conductors IConductors 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:

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 21.2bConcepTest 21.2b Conductors IIConductors II1) 0 0

2) + –

3) – +

4) + +

5) – –

0 0

? ?

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

Coulomb’s law:

This equation gives the magnitude of the force between two charges.

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.

Coulomb’s Law

Charge on the electron:

e = 1.602 x 10-19 C.

Electric charge is quantized in units of the electron charge.

Coulomb’s Law

The proportionality constant k can also be written in terms of ε0, the permittivity of free space:

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.

Coulomb’s Law

QQ QQF1 = 3 N F2 = ?

1) 1.0 N

2) 1.5 N

3) 2.0 N

4) 3.0 N

5) 6.0 N

What is the magnitude What is the magnitude

of the force of the force FF22??

Coulomb’s Law ICoulomb’s Law I

In its vector form, the Coulomb force is:

Coulomb’s Law

Coulomb’s Law

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?

Coulomb’s LawThree 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:

3R

+Q –– 4Q

Two balls with charges Two balls with charges ++QQ and and ––44QQ

are fixed at a separation distance are fixed at a separation distance

of of 33RR. Is it possible to place . Is it possible to place

another charged ball another charged ball QQ00 anywhereanywhere

on the line such that the net force on the line such that the net force

on on QQ00 will be zero? will be zero?

Electric Force IIIElectric Force III

1) yes, but only if QQ00 is positiveis positive

2) yes, but only if QQ00 is negativeis negative

3) yes, independent of the sign

(or value) of QQ00

4) no, the net force can never

be zero

Coulomb’s LawElectric force using vector components.

Calculate the net electrostatic force on charge Q3 shown in the figure due to the charges Q1 and Q2.

Solution:

Solution:

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

Forces in 2DForces in 2D

Coulomb’s Law

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?

The electric field is defined as the force on a small charge, divided by the magnitude of the charge:

The Electric Field

The Electric Field

An electric field surrounds every charge.

For a point charge:

The Electric Field

Force on a point charge in an electric field:

The Electric Field

The Electric FieldA 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 10-16 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.

The Electric FieldElectric 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.

Solution:

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

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

43

2 1

-2 C

-2 C

5) E = 0

Superposition ISuperposition I

43

2 1

-2 C

-2 C -2 C

-2 C

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

What is the direction of

the electric field at the

position of the X ?

4

32

1

+Q

-Q +Q

5

Superposition IIISuperposition III

The Electric FieldE 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)?

Solution:

Solution:

The Electric Field

E 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:

Solution:

Solution:

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!

The Electric Field

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.

Continuous Charge Distributions

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.

Continuous Charge Distributions

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:

Continuous Charge Distributions

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:The components of E parallel to the wire must add to zero by symmetry.

Continuous Charge Distributions

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, for a ring with a radius r, we know its contribution to the electric field

r

z

θdr

Solution:

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.

Infinite plane

Continuous Charge Distributions

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.

Field Lines

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

Proton and Electron IIIProton and Electron III

1) charges are equal and positive

2) charges are equal and negative

3) charges are equal and opposite

4) charges are equal, but sign is

undetermined

5) charges cannot be equal

Q2Q1 x

y

E

Two charges are fixed along Two charges are fixed along

the the xx axis. They produce an axis. They produce an

electric field electric field EE directed along directed along

the negative the negative yy axis at the axis at the

indicated point. Which of indicated point. Which of

the following is true?the following is true?

Find the ChargesFind the Charges

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

Electric dipole: two equal charges, opposite in sign:

Field Lines

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

Field Lines

uniform field

Q

In a uniform electric field in empty In a uniform electric field in empty

space, a 4 C charge is placed and it space, a 4 C charge is placed and it

feels an electric force of 12 N. If this feels an electric force of 12 N. If this

charge is removed and a 6 C charge charge is removed and a 6 C charge

is placed at that point instead, what is placed at that point instead, what

force will it feel?force will it feel?

1) 12 N

2) 8 N

3) 24 N

4) no force

5) 18 N

Uniform Electric FieldUniform Electric Field

Electric Field Lines IElectric Field Lines I

What are the signs of the

charges whose electric

fields are shown at right?

1)

2)

3)

4)

5) no way to tell

Electric Field Lines IIElectric Field Lines II

Which of the charges has

the greater magnitude?

1)

2)

3) both the same

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

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.

Conductors

The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.

Conductors

Conductors

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?

Motion 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.

Motion in an Electric FieldElectron 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.

Solution:

Motion in an Electric Field

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:

Electric DipolesAn 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.

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

Electric DipolesThe 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:

lrr

pE

lr

pE

,4

1

)4(4

1

30

23220

Electric DipolesDipole 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?

Solution:

• 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

• 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

• 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.

Summary

Gauss’s Law

• Electric Flux

• Gauss’s Law

• Applications of Gauss’s Law

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

Electric flux:

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

Electric Flux

Electric FluxElectric 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°.

Solution:

Flux through a closed surface:

Electric Flux

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.

Gauss’s Law

Gauss’s LawFor a point charge,

Therefore,

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

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.

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.

Gauss’s LawFlux 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?

Applications of Gauss’s LawSpherical 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?

Solution:

Applications of Gauss’s LawSolid 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).

Solution:

Applications of Gauss’s LawNonuniformly 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.

Solution:

Applications of Gauss’s Law

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.

Solution:Since the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere.

Applications of Gauss’s LawInfinite 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.

Solution:We expect E to be perpendicular to the plane.

Applications of Gauss’s LawElectric 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.

Applications of Gauss’s LawThe 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.

Applications of Gauss’s LawConductor 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?

Applications of Gauss’s LawProcedure 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.

• 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.

Summary