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PHYS 1444Lecture #3

• Chapter 21: E-field examples • Chapter 22: Gauss’ Law Examples• Chapter 23: Potential

- Electric Potential Energy- Electric Potential

Tuesday, June 12, 2012Ryan Hall

for Dr. Andrew Brandt

Homework on Ch 21-22 due today at midnightHomework #3 (Ch 23) Due next Tuesday June 19th at midnight

Tuesday June 12, 2012 PHYS 1444 Ryan Hall

Example 21-7Two 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: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x 108 N/C. b. We don’t know the relative lengths of E1 and E2 until we do the calculation. 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. 2

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.

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21-7 Electric Field for Continuous Charge DistributionsExample 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).

Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)

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Example 21-8Calculate 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.

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21-7 Electric Field Calculations for Continuous Charge Distributions

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.

Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its equilibrium position, it will accelerate away from the ring. 6

21-7 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/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 we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.

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Example 22 – 2 Flux from Gauss’ Law: Consider the two Gaussian surfaces, A1 and A2, shown in the figure. 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?

• The surface A1 encloses the charge +Q, so from Gauss’ law we obtain the total net flux

• For A2 the charge, +Q, is outside the surface, so the total net flux is 0.

E dA rr

Ñ

E dA rr

Ñ

0

Q

0

00

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Example 22 – 5 Long uniform line of charge: A very long straight wire possesses a uniform positive charge per unit length, . Calculate the electric field at points near but outside the wire, far from the ends.• Which direction do you think the field due to the charge on the wire is?

– Radially outward from the wire, the direction of radial vector r.• Due to cylindrical symmetry, the field is constant anywhere on the

Gaussian surface of a cylinder that surrounds the wire.– The end surfaces do not contribute to the flux at all. Why?

• Because the field vector E is perpendicular to the surface vector dA.

• From Gauss’ lawE dA

rrÑ

02E

r

Solving for E

E dAÑ 2E rl 0

enclQ

0

l

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).

Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r0

3 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r0

3).

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Electric Potential Energy• Concept of energy is very useful in solving mechanical problems• Conservation of energy makes solving complex problems easier. • When can the potential energy be defined?

– Only for a conservative force.– The work done by a conservative force is independent of the path. What

does it only depend on?? • The difference between the initial and final positions

– Can you give me an example of a conservative force?• Gravitational force

• Is the electrostatic force between two charges a conservative force?– Yes. Why?– The dependence of the force on distance is identical to that of the

gravitational force.• The only thing matters is the direct linear distance between the object not the

path.

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Electric Potential Energy• What is the definition of change in electric potential energy Ub –Ua?

– The gain (or loss) of potential energy as the charge moves from point a to point b.– The negative work done on the charge by the electric force to move it from a to b.

• Let’s consider an electric field between two parallel plates w/ equal but opposite charges– The field between the plates is uniform since the gap is

small and the plates are infinitely long…

• What happens when we place a small charge, +q, on a point at the positive plate and let go?– The electric force will accelerate the charge toward

negative plate. – What kind of energy does the charged particle gain?

• Kinetic energy

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Electric Potential Energy• What does this mean in terms of energies?

– The electric force is a conservative force.– Thus, the mechanical energy (K+U) is conserved

under this force.– The charged object has only electric potential

energy at the positive plate.– The electric potential energy decreases and – Turns into kinetic energy as the electric force works

on the charged object and the charged object gains speed.

• Point of greatest potential energy for– Positively charged object– Negatively charged object

PE=KE=ME=

U 00 KU

U+K=constK

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Electric Potential• How is the electric field defined?

– Electric force per unit charge: F/q• We can define electric potential (potential) as

– The electric potential energy per unit charge– This is like the voltage of a battery…

• Electric potential is written with a symbol V– If a positive test charge q has potential energy Ua at

a point a, the electric potential of the charge at that point is a

a

UV

q

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Electric Potential• Since only the difference in potential energy is meaningful,

only the potential difference between two points is measurable

• What happens when the electric force does “positive work”?– The charge gains kinetic energy– Electric potential energy of the charge decreases

• Thus the difference in potential energy is the same as the negative of the work, Wba, done on the charge by the electric field to move the charge from point a to b.

• The potential difference Vba is

baV b aV V b aU U

q

baW

q

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• What does the electric potential depend on?– Other charges that create an electric field– What about the test charge?

• No, the electric potential is independent of the test charge• Test charge gains potential energy by existing in the potential created by other charges

• Which plate is at a higher potential?– Positive plate. Why?

• Since positive charge has the greatest potential energy.– What happens to the positive charge if it is let go?

• It moves from higher potential to lower potential– How about a negative charge?

• Its potential energy is higher on the negative plate. Thus, it moves from negative plate to positive. Potential difference is the same for

a negative charge at the negative plate as a positive charge at the positive plate.

• The unit of the electric potential is Volt (V).• From the definition, 1V = 1J/C.

Zero point of electric potential can be chosen arbitrarily.

Often the ground, a conductor connected to Earth is zero.

A Few Things about Electric Potential

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Example 23 – 1 A negative charge: Suppose a negative charge, such as an electron, is placed at point b in the figure. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?• An electron placed at point b will move toward the positive plate

since it was released at its highest potential energy point.• It will gain kinetic energy as it moves toward left, decreasing its

potential energy.• Note the electron is moving from point b at a lower potential to

point a at a higher potential. V=Va-Vb>0.• This is because the potential is generated by the charges on the

plates not by the electron.

e-

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Electric Potential and Potential Energy • What is the definition of the electric potential?

– The potential energy difference per unit charge• How would you express the potential energy that a charge q

would obtain when it is moved between point a and b with the potential difference Vba?

– In other words, if an object with charge q moves through a potential difference Vba, its potential energy changes by qVba.

• So electric potential is– A measure of how much energy an electric charge can acquire in a

given situation– related to how much work a given charge can do.

b aU U b aq V V baqV

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Comparisons of Potential Energies • Let’s compare gravitational and electric potential energies

2mm

• What is the potential energy of each rock?– mgh and 2mgh

• Which rock has a bigger potential energy?– The rock with a larger mass

• Why?– It’s got a bigger mass.

• What is the potential energy of each charge?– +QVba and +2QVba

• Which object has a bigger potential energy?– The object with a larger charge.

• Why?– It’s got a bigger charge.

The “potential” is the same but the heavier rock or larger charge can do a greater work.

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Some Typical Voltages Sources Approximate Voltage

Thundercloud to ground 108 VHigh-Voltage Power Lines 106 VPower supply for TV tube 104 VAutomobile ignition 104 VHousehold outlet 102 VAutomobile battery 12 VFlashlight battery 1.5 VResting potential across nerve membrane 10-1 VPotential changes on skin (EKG and EEG) 10-4 V

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Electric Potential and Potential Energy • The electric potential difference gives potential energy (or

the possibility to do work) based on the charge of the object.• So what is happening in batteries or generators?

– They maintain a potential difference.– The actual amount of energy used or transformed depends on how

much charge flows.– How much is the potential difference maintained by a car’s battery?

• 12Volts

– If for a given period, 5C charge flows through the headlight lamp, what is the total energy transformed?

• Etot=5C*12V=60 What is the unit?

– If it is left on twice as long? Etot=10C*12V=120J.C*J/C=J (Joules)

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Example 23 – 2 Electrons in a TV tube: Suppose an electron in the picture tube of a television set is accelerated from rest through a potential difference Vba=+5000V. (a) What is the change in potential energy of the electron? (b) What is the speed of the electron (m=9.1x10-31kg) as a result of this acceleration? (c) Repeat for a proton (m=1.67x10-27kg) that accelerates through a potential difference of Vba=-5000V.

• (a) What is the charge of an electron?–

• So what is the change of its potential energy?

U baqV 19 161.6 10 5000 8.0 10baeV C V J

191.6 10e C

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Example 23 – 2 • (b) Speed of the electron?

– The entire potential energy of the electron is transformed into kinetic energy. Thus the equation is

ev

K

• (c) Speed of a proton that accelerates through V=-5000V?

pv

K

210

2 e em v W U ( )baU eV

16 168.0 10 8.0 10J J

2

e

K

m

167

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2 8.0 104.2 10 /

9.1 10m s

210

2 p pm v 168.0 10ba baW U e V eV J

2 ba

p

eV

m

165

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2 8.0 109.8 10 /

1.67 10m s

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Electric Potential and Electric Field• The effect of a charge distribution can be

described in terms of electric field or electric potential.– What kind of quantities are the electric field and the

electric potential?• Electric Field:• Electric Potential:

– Since electric potential is a scalar quantity, it often can make problem solving easier.

VectorScalar

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Electric Potential and Electric Field• The potential energy is an (independent of path) function

expressed in terms of a (conservative) force.

• The potential difference is the potential energy difference per unit charge

– This formula can be used to determine Vba when the electric field is given.

• When the field is uniform

b aU U

baV

b aV V baV Edor

Unit of the electric field in terms of potential? V/m Can you derive this from N/C?

b aV V b aU U

q

b

a

Fdl

q

rr b

aE dl

rr

b

aE dl

rr b

aE dl Ed

b

aF dl

rr

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50V

5cm

Example 23 – 3 Uniform electric field obtained from voltage: Two parallel plates are charged to a voltage of 50 V. If the separation between the plates is 5.0 cm, calculate the magnitude of the electric field between them, ignoring any fringe effects.

E V

d

50

5.0

V

cm

What is the relationship between electric field and the potential for a uniform field? V Ed

Solving for E 2

50

5 10

V

m

1000 /V m

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Electric Potential due to Point Charges• What is the electric field due to a point charge Q at a

distance r?

• Electric potential due to the field E for moving from point ra to rb away from the charge Q is

b aV V b

a

r

rE dl

rr

E 20

1

4

Q

r

2

Qkr

20

ˆˆ

4

b

a

r

r

Q rrdr

r

20

1

4

b

a

r

r

Qdr

r

0

1 1

4 b a

Q

r r

Notice how the integral is carried out in the radial direction.

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Electric Potential due to Point Charges• Since only the differences in potential have physical

meaning, we can choose at .• The electrical potential V at a distance r from a

single point charge is

• So the absolute potential from a single point charge depends only on the magnitude of the point charge and the distance from it

V 0

1

4

Q

r

0bV br

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• What are the differences between the electric potential and the electric field?– Electric potential

• Electric potential energy per unit charge• Inversely proportional to the distance• Simply add the potential from each of the charges to obtain the total potential

from multiple charges, since potential is a scalar quantity

– Electric field• Electric force per unit charge• Inversely proportional to the square of the distance• Need vector sums to obtain the total field from multiple charges

• Potential for a positive charge is large near the charge and decreases to 0 at large distances.

• Potential for the negative charge is small (large magnitude but negative) near the charge and increases with distance to 0

Properties of the Electric Potential

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1

4

QE

r

r

0

1

4

QV

r

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Shape of the Electric Potential• So, how does the electric potential look like as a function of

distance?– What is the formula for the potential by a single charge?

V 0

1

4

Q

rPositive Charge Negative Charge

A uniformly charged sphere would have the same potential as a single point charge.

What does this mean? Uniformly charged sphere behaves like all the charge is on the single point in the center.

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Since we obtain

Example 23 – 6Work to bring two positive charges close together: What minimum work is required by an external force to bring a charge q=3.00 μC from a great distance away ( ) to a point 0.500 m from a charge Q=20.0 μC?What is the work done by the electric field in terms of potential energy and potential?

W

0.500 ,b ar m r

W

In other words, the external force must input work of +1.08J to bring the charge 3.00C from infinity to 0.500m from the 20.0C charge.

baqV 04 b a

q Q Q

r r

0

04 b

q Q

r

04 b

q Q

r 9 2 2 6 68.99 10 3.00 10 20.00 10

1.080.500

N m C C CJ

m

r