ELECTRIC POTENTIAL

AP Physics: Electricity & Magnetism

POTENTIAL ENERGY

What is potential energy?

The ability to do work

The energy possessed by an object due to its

position in a force field.

WHAT IS WORK?

Work is an energy transfer

A constant force applied over a distance in the same

direction.

W F d

WORK AND POTENTIAL ENERGY

W Fsi

sf ds UU i U f

The work done by a conservative force equals the decrease in the potential

energy of the particle.

WORK DONE BY AN ELECTRIC FIELD

W FcA

B ds U

W q0EA

B ds U

U q0 EA

B ds

This is called a line or path integral which is independent of the path taken from point A

to B.

THE CHANGE IN POTENTIAL ENERGY PER UNIT CHARGE:

U

q0

EA

B dsV

This value is called Electric

Potential and is independent of the value of q

VU

q0

The potential at any point in the

field:

ELECTRIC POTENTIAL:

Be careful, electric potential is not potential energy.

The work per unit charge that an external agent must perform to move a test charge

from A to B without a change in kinetic energy

Units? J/C

POTENTIAL AT A POINT P

VP E

P ds

The electric field at an arbitrary point equals the work required per unit charge

to bring a positive test charge from infinity to that point

A NOTE ON UNITS:

J

C

N

Cm

V

U

q0

EA

B dsV

Joule

Coulomb

Newton meter

Coulomb

Volt

THE ELECTRON VOLT

The energy that an electron or proton gains or loses by moving through a potential difference

of 1V

1.60210 19 J 1eV

POTENTIAL DIFFERENCES IN A UNIFORM ELECTRIC FIELD:

What is the potential difference between two points if the displacement is parallel to the field

lines?

VVB VA E dsA

B

E cos0dsA

B EdsA

B E dsA

B

V Ed

POTENTIAL AT POINTS B & C

E

A

B

C

VB VC

Cross-section of equipotential

surface

EQUIPOTENTIAL SURFACE

The name given to any surface consisting of a

continuous distribution of points having the same

electric potential.

No work is done in moving a test charge between any two points on an equipotential

surface.

EQUIPOTENTIAL SURFACE Red lines are electric field lines Blue lines are equipotential surfaces Equipotential surfaces must be

perpendicular to field lines.

EQUIPOTENTIAL SURFACE

EQUIPOTENTIAL SURFACE

GAIN OR LOSS?A negative charge gains electric

potential energy when it moves in the direction of the electric field. Explain

this.A proton loses potential energy when moving in the direction of the electric field, but picks up an equal amount of

kinetic energy.Work done by a field is positive when energy is given to an object from the

field (object is lowered in a gravitational field or moved in the direction of E filed lines). This occurs when Uf <Ui so, W=-

ΔU

THE GAP BETWEEN ELECTRODES IN A SPARK PLUG IS 0.06CM, TO PRODUCE AN ELECTRIC SPARK IN A GASOLINE AIR MIXTURE, AN E FIELD OF 3X106V/M MUST BE ACHIEVED. WHEN STARTING THE CAR, WHAT MINIMUM POTENTIAL DIFFERENCE MUST BE SUPPLIED BY THE IGNITION CIRCUIT?

d0.0006m

E 3106V /m

VEd

VEd

V 3106V /m 0.0006m

V1.8kV

CALCULATE THE SPEED OF A PROTON THAT IS ACCELERATED FROM REST THROUGH A POTENTIAL DIFFERENCE OF 120V. AN ELECTRON?

V120V

v0 0

v?

mp 1.6710 27kg

me 9.1110 31kg

e1.602 19 C

VU

q

qVUKE 1

2mv2

v2qV

m

vp 1.5105m/s

ve 6.5106m/s

AN ION ACCELERATED THROUGH A POTENTIAL DIFFERENCE OF 115V EXPERIENCES AN INCREASE IN KINETIC ENERGY OF 7.37X10-17 J.CALCULATE THE CHARGE OF THE ION

V115V

KE 7.3710 17 J

q?

qVUKE

qKE

V

q6.410 19C

HOW MUCH WORK IS DONE IN MOVING AVOGADRO'S NUMBER OF ELECTRONS FROM AN INITIAL POINT WHERE THE ELECTRIC POTENTIAL IS 9V TO A POINT WHERE THE POTENTIAL IS -5V?

VA 9V

VB 5V

Na 6.021023electrons

W ?

VVA VB

V9V ( 5V)

V14V

W UqV

qNae(6.021023)( 1.60210 19C)

q 96440.4C

W qV( 96440.4C)(14V)

W 1.35MJ

POTENTIAL ENERGY DUE TO A POINT CHARGE?

What is the electric field due to a point charge (q)?

E dAQ in

0

EAQ in

0

Field is radially outward (or inward), and the Gaussian surface is a sphere chosen to match symmetry of field.

E Q in

A0

q

4r20

kq

r2

POTENTIAL ENERGY DUE TO A POINT CHARGE?

E kq

r2

VVB VA E dsrA

rB

The following simplification can be made because the potential difference is not dependant upon the path taken through the field, only the endpoints.

V EdrrA

rB

Vkq

r2 drrA

rB

V kq1

r2 drrA

rB

V kq 1

r

rA

rB

POTENTIAL ENERGY DUE TO A POINT CHARGE?

If we want to know the potential at one particular location near a point charge we chose the first point to

be infinity.

V kq 1

r

rA

rB

Vkq1

r

rB

Vkq1

r

1

Vkq

r

SUPERPOSITION PRINCIPLE

Vkq

r

Vkqi

rii

Potential at a point due to many point charges.

This is a scalar value so it is much easier to calculate than E

field.

POTENTIAL ENERGY DUE TO MANY CHARGES.

Uq2V1 kq1q2

r12

The potential energy of a pair of charged particles separated by a distance r12. this is also the work done to move q2 from

infinity to point P where q1 is located. V1 is the electric potential at

point P due to charge q1.

Notice that if the charges are like then positive work is done to bring the charges together (they

have more energy together because they will repel). Energy delivered to the system requires

positive work.

FOR MULTIPLE CHARGES:

Ukq1q2

r12

q1q3

r13

q2q3

r23

Imagine q1 is at a fixed position. We want to bring q2 and q3 from infinity to a position near q1.

Each term represents the the work required to bring each charge to a location near the other. It does not matter which

order we bring them in from infinity.

q1 q2

q3

r12

r23

r13

YOU KNOW POTENTIAL AND YOU WANT TO KNOW ELECTRIC FIELD…

V EA

B ds

How can we use this to solve for E field?

dV E ds

If the E field only has one component…

dV E xdx

E x dV

dx

YOU KNOW POTENTIAL AND YOU WANT TO KNOW ELECTRIC FIELD…

Remember that the potential is zero for displacements perpendicular to the field

If your potential is a function of all three spatial coordinates (x, y, z)

E x dV

dx

E y dV

dy

E z dV

dz

V4x2y y2 2yz

To find the electric field, we must take the partial derivative of this potential…

E x Vxx

4x2y y2 2yz

E x x

4x2y

4yx

x2

4y2x8xy

Partial derivative: Treat other variables as constant while differentiating with respect to one variable.

V4x2y y2 2yz

E y Vyy

4x2y y2 2yz

E y 4x2 2y 2z

E z Vzz

2yz

E z 2y

E x,y,z 8xy i 4x2 2y 2z j 2y ˆ k

In vector notation this is often written as…

E Vx i

y

j

z k

is called the gradient operator

If the potential is constant in some region, what is the electric field?

If the electric field is zero in some region, what is the electric potential?

QUESTIONIf the potential is constant in some region,

what is the electric field?If the electric field is zero in some region,

what is the electric potential?

vconstant, E 0

E 0, vconstant

The difference in electric potential (voltage) measured when moving from point A to point B is equal to the work which would have to be done, per unit charge, against

the electric field to move the charge from A to B.

ELECTRIC FIELD DUE TO CONTINUOUS CHARGE DISTRIBUTIONS.

The electric potential at point P due to a continuous charge distribution can be

calculated by dividing the charged body into segments of charge dq and summing the potential contributions over all segments.

Vkq

r

Vkdq

r

POTENTIAL DUE TO A UNIFORMLY CHARGED RING:

P

a

x

dq

Vkdq

r

r a2 x2

All segments of charge are equidistant from point P

Vk

a2 x2dq

VkQ

a2 x2

POTENTIAL DUE TO A UNIFORMLY CHARGED RING AT THE CENTER:

P

a

dqThe field is zero here, therefore the potential

must be constant:

VkQ

a

this potential is the work necessary to bring a test charge from infinity to the location in

the center of the ring.

POTENTIAL OF A CHARGED CONDUCTOR

A

B

This oddly shaped conductor with an

excess positive charge is in equilibrium

Along this surface path E is always

perpendicular to the displacement ds.

E ds0

VB VA E dsA

B 0

POTENTIAL OF A CHARGED CONDUCTOR

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

Furthermore, since the electric field is zero inside the conductor, we conclude that the potential is constant everywhere inside the conductor and

equal to its value at the surface.

A

B

The electric field is large near points having small convex radii of curvature and reaches

very high values at sharp points.

A CAVITY WITHIN A CONDUCTOR

AB

A cavity in a conductor with no charge in it…The electric field inside the cavity must be

zero, regardless of the charge distribution on the outside surface of the conductor.

Because the potential on the surface of the cavity

is an equipotential surface.

ELECTRIC SHIELDING Sensitive circuits and people can be

protected if placed in a cavity inside of a conductor.

CORONA DISCHARGE:

Air can become a conductor as a result of the ionization or air molecules in regions

of high electric fields. At STP this happens around 3x106

V/m

PRACTICE QUIZ 1

.A .B

.C

.D

+2Q

-Q-Q2a

Three charges are arranged in an equilateral triangle, as shown. At which of

these points (a,b,c bisect sides, d is equidistant from other points) is the

electric potential smallest?

.E

PRACTICE QUIZ 1

.A .B

.C

.D

+2Q

-Q-Q2a

C) A small positive test charge will move towards an area of low potential. Ask

yourself “Where would an small positive charge end up if released near these

charges?”

.E

PRACTICE QUIZ 2

10V

The diagram shows a set of equipotential surfaces. At point P, what is the direction

of the electric field?

40V 70V

.P

A) left B) right

C) up the slide D) down the slideE) either left or right, which one cannot be

determined

PRACTICE QUIZ 2

10V

A) A small positive charge placed at P would move to a location of low potential (left). The force that moves it is caused by an electric field which will be in the

same direction or opposite it. Equipotential surfaces are always

perpendicular to electric field lines.

40V 70V

.P

PRACTICE QUIZ 3

What is the electric potential at a point halfway between the two

charges?

a

QQ

A) kQ/a

B) 2kQ/a

C) zero

D) 4kQ/a

E) 8kQ/a

D) Electric potential is a

scalar

VkQa

2

kQa

2

2kQ

a

2

4kQ

a

PRACTICE QUIZ 4

A solid conducting sphere carries a charge +Q. Which of the following

is true of the electric field E and the electric potential V inside the

sphere?A) E=0 and V=0

B) E=0 and V≠0

C) E≠0 and V=0

D) E≠0 and V≠0

E) It cannot be determined without knowing the radius of the sphere.

B) The field in a conductor is always zero. E is the derivative of V, so to be

zero, V must have been a constant.

E dV

dr

PRACTICE QUIZ 5A negatively charged rod is brought near a metal

object on an insulating stand, as shown. When charges stop moving, the left side of the object has an excess of positive charge, and the right

side of the object, where the radius of curvature is less, has an excess of negative charge. Which

of the following best describes the electric potential on the metal object?

- - - - - - - ---

+++ +

A) It is greatest on the + sideB) It is greatest on the – sideC) It is greatest at the centerD) It is the same everywhere on the objectE) It cannot be determined from the       information given

PRACTICE QUIZ 5

- - - - - - - ---

+++ +

A) It is greatest on the + sideB) It is greatest on the – sideC) It is greatest at the centerD) It is the same everywhere on the objectE) It cannot be determined from the       information given

D) It is the same everywhere on the object. The surface of a conductor is an equipotential surface. The E field is 0

inside the conductor and V is constant.

PRACTICE QUIZ 6

L

Four identical charged particles, each with charge Q, are fixed in place in the shape of an equilateral pyramid with sides of length L, as shown above.

What is the potential energy of this arrangement of charges?

.Q

.Q

.Q

.Q

L

L

L

L

L

PRACTICE QUIZ 6

L

.Q

.Q

.Q

.Q

L

L

L

L

L

Each pair of charges is separated by a distance L.

Each pair contributes:

There are 6 pairings:

UQ2

40L

U6Q2

40L

Uq1q2

40rIn general:

PRACTICE QUIZ 7 What is the amount of work required to

assemble 4 identical point charges of magnitude Q at the corners of a square of side s?

s

sQ Q

Q Q

2s

U kQ2 1

s

1

s

1

s

1

s

1

2s

1

2s

U kQ2 4

s

2

2s

UkQ2

s4

2

2

U5.41kQ2

s

PRACTICE QUIZ 8

A wire that has a uniform linear charge density λ is bent into the shape shown

below. Find the electric potential at point O.

2R

2R

R.O

PRACTICE QUIZ 9

A positron is given an initial velocity of 6x106m/s. It travels into a uniform electric field, directed to the left. As the positron enters the field, its electric potential is

zero. What is the electric potential at the point where the positron has a speed of

1x106m/s?

E+vi