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Quiz on the math needed today at is the result of this integral: B A dx x 2 1 )? ( is what , ) ( ) ( and known is ) ( If x z dx x z x y x y )? ( is what , ) ( ) ( and known is ) ( If integral line r z s r z r r d y y B A x dx x B A B A 1 1 1 1 2 dx x dy x z ) ( ) ( k j i r r z ˆ ˆ ˆ y x x x system coordinate Cartesian a In ), ( ) (
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Quiz on the math needed today What is the result of this integral:

Jan 12, 2016

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Page 1: Quiz on the math needed today What is the result of this integral:

Quiz on the math needed today

What is the result of this integral: B

A

dxx 2

1

)?( is what ,)()( andknown is )( If xzdxxzxyxy

)?( is what ,)()( andknown is )( Ifintegral line

rzsrzrr

dyy

BAxdx

x

B

A

B

A

11112

dx

xdyxz

)()(

kji

rrz

ˆˆˆ

y

xxx system coordinateCartesian aIn

),()(

Page 2: Quiz on the math needed today What is the result of this integral:

Chapter 25

Electric Potential

Page 3: Quiz on the math needed today What is the result of this integral:

A review of gravitational potential

B

When object of mass m is on ground level B, we define that it has zero gravitational potential energy. When we let go of this object, if will stay in place.

When this object is moved to elevation A, we say that it has gravitational potential energy mgh. h is the distance from B to A. When we let go of this object, if will fall back to level B.

A

When this object is at elevation A, it has gravitational potential energy UA-UB= mgh. UA is the potential energy at point A with reference to point B. When the object falls from level A to level B, the potential energy change: ΔU =UB-UA

The gravitational force does work and causes the potential energy change: W = mgh = UA-UB= -ΔU

h

mg

We also know that the gravitational force is conservative: the work it does to the object only depends on the two levels A and B, not the path the object moves.

Page 4: Quiz on the math needed today What is the result of this integral:

Introduction of the electric potential, a special case: the electric field is a constant.

EF

0q

When a charge q0 is placed inside an electric field, it experiences a force from the field:

When the charge is released, the field moves it from A to B, doing work:

Edqq 00W dEdF

If we define the electric potential energy of the charge at point A UA and B UB, then:

UUUEdq BA 0WIf we define UB=0, then UA= q0Ed is the electric potential energy the charge has at point A. We can also say that the electric field has an electric potential at point A. When a charge is placed there, the charge acquires an electric potential energy that is the charge times this potential.

Page 5: Quiz on the math needed today What is the result of this integral:

Electric Potential Energy, the general caseWhen a charge is moved from point A to point B in an electric field, the charge’s electric potential energy inside this field is changed from UA to UB: AB UUU

EF

0qThe force on the charge is:

B

A

BA dqWUUU sE

0 force) field the(of

So we have this final formula for electric potential energy and the work the field force does to the charge:

When the motion is caused by the electric field force on the charge, the work this force does to the charge cause the change of its electric potential energy, so: UW

Page 6: Quiz on the math needed today What is the result of this integral:

Electric Potential Energy, final discussion

Electric force is conservative. The line integral does not depend on the path from A to B; it only depends on the locations of A and B.

B

A

BA dqUUU sE

0

A BLine integral paths

Page 7: Quiz on the math needed today What is the result of this integral:

The electric potential energy of a charge q0 in the field of a charge Q?

Q

q0

R

Reference point:We usually define the electric potential of a point charge to be zero (reference) at a point that is infinitely far away from the point charge.

B

A

BA dqUUU sE

0Applying this formula:

Where point A is where the charge q0 is, point B is infinitely far away.

R

Qkdr

r

Qk

drr

Qkdd

ddˆr

Qk

e

R

e

e

e

2

2

2

and

so

,

rEsE

rsrE

And the result is a scalar!

So the final answer is

R

QqkRU e

0 )(

Page 8: Quiz on the math needed today What is the result of this integral:

Electric Potential, the definition The potential energy per unit charge, U/qo, is the

electric potential The potential is a characteristic of the field only

The potential energy is a characteristic of the charge-field system

The potential is independent of the value of qo The potential has a value at every point in an electric field

The electric potential is

As in the potential energy case, electric potential also needs a reference. So it is the potential difference ΔV that matters, not the potential itself, unless a reference is specified (then it is again ΔV).

o

UV

q

Page 9: Quiz on the math needed today What is the result of this integral:

Electric Potential, the formula The potential is a scalar quantity

Since energy is a scalar As a charged particle moves in an electric

field, it will experience a change in potential

B

A

BA dVVV sE

reference) (often the

Page 10: Quiz on the math needed today What is the result of this integral:

Potential Difference in a Uniform Field

dEsEsE

B

A

B

A

BA ddVVV

The equations for electric potential can be simplified if the electric field is uniform:

BABA VV,VVV

or 0

direction, same the and i.e.,

dE 0, dE

When:

This is to say that electric field lines always point in the direction of decreasing electric potential

Page 11: Quiz on the math needed today What is the result of this integral:

Electric Potential, final discussion

The difference in potential is the meaningful quantity

We often take the value of the potential to be zero at some convenient point in the field

Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field

Page 12: Quiz on the math needed today What is the result of this integral:

Electric Potential, electric potential energy and Work

When there is electric field, there is electric potential V.

When a charge q0 is in an electric field, this charge has an electric potential energy U in this electric field: U = q0 V.

When this charge q0 is move by the electric field force, the work this field force does to this charge equals the electric potential energy change -ΔU: W = -ΔU = -q0 ΔV.

Page 13: Quiz on the math needed today What is the result of this integral:

Units The unit for electric potential energy is the unit for energy joule

(J). The unit for electric potential is volt (V):

1 V = 1 J/C This unit comes from U = q0 V (here U is electric potential

energy, V is electric potential, not the unit volt) It takes one joule of work to move a 1-coulomb charge through a

potential difference of 1 volt

But from

We also have the unit for electric potential as 1 V = 1 (N/C)mSo we have that 1 N/C (the unit of ) = 1 V/m This indicates that we can interpret the electric field as a

measure of the rate of change with position of the electric potential

B

A

dV sE

E

Page 14: Quiz on the math needed today What is the result of this integral:

Electron-Volts, another unit often used in nuclear and particle physics

Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt

One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt 1 eV = 1.60 x 10-19 J

Page 15: Quiz on the math needed today What is the result of this integral:

Direction of Electric Field, energy conservation

As pointed out before, electric field lines always point in the direction of decreasing electric potential

So when the electric field is directed downward, point B is at a lower potential than point A

When a positive test charge moves from A to B, the charge-field system loses potential energy through doing work to this charge

Where does this energy go?

PLAYACTIVE FIGURE

2502

It turns into the kinetic energy of the object (with a mass) that carries the charge q0.

Page 16: Quiz on the math needed today What is the result of this integral:

Equipotentials = equal potentials Points B and C are at a lower

potential than point A Points B and C are at the

same potential All points in a plane

perpendicular to a uniform electric field are at the same electric potential

The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential

Page 17: Quiz on the math needed today What is the result of this integral:

Charged Particle in a Uniform Field, ExampleQuestion: a positive charge (mass m) is released from rest and moves in the direction of the electric field. Find its speed at point B.

Solution: The system loses potential energy: -ΔU=UA-UB=qEd

The force and acceleration are in the direction of the field

Use energy conservation to find its speed:

m

qEdv

qEdmv

2

2

1 2

Page 18: Quiz on the math needed today What is the result of this integral:

Potential and Point Charges

A positive point charge produces a field directed radially outward

The potential difference between points A and B will be

1 1B A e

B A

V V k qr r

Page 19: Quiz on the math needed today What is the result of this integral:

Potential and Point Charges, cont.

The electric potential is independent of the path between points A and B

It is customary to choose a reference potential of V = 0 at rA = ∞

Then the potential at some point r is

e

qV k

r

Page 20: Quiz on the math needed today What is the result of this integral:

Electric Potential of a Point Charge

The electric potential in the plane around a single point charge is shown

The red line shows the 1/r nature of the potential

Page 21: Quiz on the math needed today What is the result of this integral:

Electric Potential with Multiple Charges

The electric potential due to several point charges is the sum of the potentials due to each individual charge This is another example of the superposition

principle The sum is the algebraic sum

V = 0 at r = ∞

ie

i i

qV k

r

Page 22: Quiz on the math needed today What is the result of this integral:

Immediate application: Electric Potential of a Dipole

The graph shows the potential (y-axis) of an electric dipole

The steep slope between the charges represents the strong electric field in this region

)11

(

r-rr-rqkVVV e

Work on the board to prove it.

Page 23: Quiz on the math needed today What is the result of this integral:

Potential Energy of Multiple Charges

Consider two charged particles

The potential energy of the system is

Use the active figure to move the charge and see the effect on the potential energy of the system

1 2

12e

q qU k

r

PLAYACTIVE FIGURE

2509

Page 24: Quiz on the math needed today What is the result of this integral:

More About U of Multiple Charges

If the two charges are the same sign, U is positive and external work (not the one from the field force) must be done to bring the charges together

If the two charges have opposite signs, U is negative and external work is done to keep the charges apart

Page 25: Quiz on the math needed today What is the result of this integral:

U with Multiple Charges, take 3 as an example

If there are more than two charges, then find U for each pair of charges and add them

For three charges:

The result is independent of the order of the charges

1 3 2 31 2

12 13 23e

q q q qq qU k

r r r

Page 26: Quiz on the math needed today What is the result of this integral:

Finding E From V

This is straight forward

sEdVFrom We have V

xxxV )( kjiE

If E is one dimensional (say along the x-axis)dx

dVEx

If E is only a function of r (the point charge case):

r)rE

)(( and ,)( rEdr

dVrE

Page 27: Quiz on the math needed today What is the result of this integral:

Find V for an Infinite Sheet of Charge

We know the a constant From

We have The equipotential lines are the

dashed blue lines The electric field lines are the

brown lines The equipotential lines are

everywhere perpendicular to the field lines

02

E

sEdVEdV d

Page 28: Quiz on the math needed today What is the result of this integral:

E and V for a Point Charge

The equipotential lines are the dashed blue lines

The electric field lines are the brown lines

The equipotential lines are everywhere perpendicular to the field lines

Page 29: Quiz on the math needed today What is the result of this integral:

E and V for a Dipole

The equipotential lines are the dashed blue lines

The electric field lines are the brown lines

The equipotential lines are everywhere perpendicular to the field lines

Page 30: Quiz on the math needed today What is the result of this integral:

When you use a computer (program) to calculate electric Potential for a Continuous Charge Distribution:

Consider a small charge element dq Treat it as a point charge

The potential at some point due to this charge element is

e

dqdV k

r

Page 31: Quiz on the math needed today What is the result of this integral:

V for a Continuous Charge Distribution, cont.

To find the total potential, you need to integrate to include the contributions from all the elements

This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions

e

dqV k

r

Page 32: Quiz on the math needed today What is the result of this integral:

V for a Uniformly Charged Ring

P is located on the perpendicular central axis of the uniformly charged ring The ring has a radius a

and a total charge Q

2 2

ee

k QdqV k

r a x

Page 33: Quiz on the math needed today What is the result of this integral:

V for a Uniformly Charged Disk

The ring has a radius R and surface charge density of σ

P is along the perpendicular central axis of the disk

1

2 2 22 eV πk σ R x x

Page 34: Quiz on the math needed today What is the result of this integral:

V for a Finite Line of Charge

A rod of line ℓ has a total charge of Q and a linear charge density of λ

2 2

lnek Q aV

a

Page 35: Quiz on the math needed today What is the result of this integral:

Prove that V is everywhere the same on a charged conductor in equilibrium

Inside the conductor, because is 0, , so ΔV=0

On the surface, consider two points on the surface of the charged conductor as shown

is always perpendicular to the displacement

Therefore, Therefore, the potential

difference between A and B is also zero

E

ds

0d E s

E

0d E s

Page 36: Quiz on the math needed today What is the result of this integral:

Summarize on potential V of a charged conductor in equilibrium

V is constant everywhere on the surface of a charged conductor in equilibrium ΔV = 0 between any two points on the surface

The surface of any charged conductor in electrostatic equilibrium is an equipotential surface

Because the electric field is zero inside the conductor, we conclude that the electric potential is constant everywhere inside the conductor and equal to the value at the surface

Page 37: Quiz on the math needed today What is the result of this integral:

E Compared to V

The electric potential is a function of r

The electric field is a function of r2

The effect of a charge on the space surrounding it: The charge sets up a

vector electric field which is related to the force

The charge sets up a scalar potential which is related to the energy

Page 38: Quiz on the math needed today What is the result of this integral:

Cavity in a Conductor

Assume an irregularly shaped cavity is inside a conductor

No charges are inside the cavity

The electric field inside the conductor must be zero (can you prove that?)

Page 39: Quiz on the math needed today What is the result of this integral:

Cavity in a Conductor, cont

The electric field inside does not depend on the charge distribution on the outside surface of the conductor

For all paths between A and B,

A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity

0B

B A AV V d E s

Page 40: Quiz on the math needed today What is the result of this integral:

Millikan Oil-Drop Experiment – Experimental Set-Up

PLAYACTIVE FIGURE

Page 41: Quiz on the math needed today What is the result of this integral:

Millikan Oil-Drop Experiment

Robert Millikan measured e, the magnitude of the elementary charge on the electron

He also demonstrated the quantized nature of this charge

Oil droplets pass through a small hole and are illuminated by a light

Page 42: Quiz on the math needed today What is the result of this integral:

Oil-Drop Experiment, 2

With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electron

The drop reaches terminal velocity with

D mF g

Page 43: Quiz on the math needed today What is the result of this integral:

Oil-Drop Experiment, 3

When an electric field is set up between the plates The upper plate has a

higher potential The drop reaches a

new terminal velocity when the electrical force equals the sum of the drag force and gravity

Page 44: Quiz on the math needed today What is the result of this integral:

Oil-Drop Experiment, final

The drop can be raised and allowed to fall numerous times by turning the electric field on and off

After many experiments, Millikan determined: q = ne where n = 0, -1, -2, -3, … e = 1.60 x 10-19 C

This yields conclusive evidence that charge is quantized

Use the active figure to conduct a version of the experiment

Page 45: Quiz on the math needed today What is the result of this integral:

Van de Graaff Generator Charge is delivered continuously to a

high-potential electrode by means of a moving belt of insulating material

The high-voltage electrode is a hollow metal dome mounted on an insulated column

Large potentials can be developed by repeated trips of the belt

Protons accelerated through such large potentials receive enough energy to initiate nuclear reactions

Page 46: Quiz on the math needed today What is the result of this integral:

Electrostatic Precipitator An application of electrical discharge in gases

is the electrostatic precipitator It removes particulate matter from

combustible gases The air to be cleaned enters the duct and

moves near the wire As the electrons and negative ions created by

the discharge are accelerated toward the outer wall by the electric field, the dirt particles become charged

Most of the dirt particles are negatively charged and are drawn to the walls by the electric field