Electric Potential A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007.

Electric PotentialElectric PotentialA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: Objectives: After completing After completing this module, you should be this module, you should be

able to:able to:• Understand an apply the concepts of electric potential

energy, electric potential, and electric potential difference.

• Calculate the work required to move a known charge from one point to another in an electric field created by point charges.

• Write and apply relationships between the electric field, potential difference, and plate separation for parallel plates of equal and opposite charge.

Review: Work and EnergyReview: Work and Energy

Work is defined as the product of displacement d and a parallel applied force F.

Work is defined as the product of displacement d and a parallel applied force F.

Work = Fd; Units: 1 J = 1 N mWork = Fd; Units: 1 J = 1 N m

Potential Energy U is defined as the ability to do work by virtue of position or condition. (Joules)

Potential Energy U is defined as the ability to do work by virtue of position or condition. (Joules)

Kinetic Energy K is defined as the ability to do work by virtue of motion (velocity). (Also in joules)

Kinetic Energy K is defined as the ability to do work by virtue of motion (velocity). (Also in joules)

Signs for Work and EnergySigns for Work and Energy

Work (Fd) is positive if an applied force F is in the same direction as the displacement d.

Work (Fd) is positive if an applied force F is in the same direction as the displacement d.

A

B

mF

mg

d

The force The force FF does does positive work. positive work.The force The force mgmg does does negative worknegative work.

The The P.E.P.E. at at BB relative to relative to AA is is positivepositive because the field can do because the field can do positive work ifpositive work if mm is released. is released.

P.E.P.E. at at AA relative to relative to BB is is negativenegative; ; outside force needed to move outside force needed to move mm..

Gravitational Work and Gravitational Work and EnergyEnergy

Consider work against g to move m from A to B, a vertical height h.

A

B

hmF

Work = Fh = mgh

At level B, the potential energy U is:

U = mgh (gravitational)

The external force does positive work; the gravity g does negative work.

The external force F against the g-field increases the potential energy. If released the field gives work back.

The external force F against the g-field increases the potential energy. If released the field gives work back.

gmg

Electrical Work and Electrical Work and EnergyEnergy

An external force F moves +q from A to B against the field force qE.

Work = Fd = (qE)dAt level B, the potential energy U is:

U = qEd (Electrical)

The E-field does negative work; External force does positive work.

The external force F against the E-field increases the potential energy. If released the field gives work back.

The external force F against the E-field increases the potential energy. If released the field gives work back.

B + + + +

- - - -A

++q d

qE E

Fe

Work and Negative Work and Negative ChargesCharges

Suppose a negative charge –q is moved against E from A to B.

Work by E = qEd

At A, the potential energy U is:

U = qEd (Electrical)

No external force is required !

B + + + +

- - - -A

E

dqE-q

The E-field does positive work on –q decreasing the potential energy. If released from B nothing happens.

The E-field does positive work on –q decreasing the potential energy. If released from B nothing happens.

Work to Move a ChargeWork to Move a Charge

++++ +

+++Q

qE

F

Work to move +q from Ato

B.

ra

rb

2aa

kqQF

r

avga b

kqQF

r r

At A:

At B:

Avg. Force:

2bb

kqQF

r

Distance: ra - rb

( )a ba b

kQqWork Fd r r

r r 1 1

b a

Work kQqr r

1 1

b a

Work kQqr r

Absolute Potential EnergyAbsolute Potential Energy

++++ +

+++Q

qE

F

ra

rb

1 1

b a

Work kQqr r

1 1

b a

Work kQqr r

Absolute P.E. is Absolute P.E. is relative to relative to

It is work to bring It is work to bring +q+q from infinity to from infinity to point near point near QQ—i.e., —i.e., from from to rto rbb

1 1

b b

kQqWork kQq

r r

Absolute Potential Energy:

kQqU

r

0

Example 1.Example 1. What is the potential What is the potential energy if a energy if a +2 nC+2 nC charge moves from charge moves from to point to point AA, , 8 cm8 cm away from a away from a +6 +6 CC charge?charge?

+6 C

+Q

A

+2 nC

kQqU

rPotential Potential

Energy:Energy:2

2

9 -6 -9NmC

(9 x 10 )( 6 x 10 C)(+2 x 10 C)

(0.08 m)U

The The P.E.P.E. will be will be positivepositive at at point point AA, because the , because the field field can can do do ++ work if work if qq is released. is released.

U = 1.35 mJU = 1.35 mJ Positive potential energy

8 cm

Signs for Potential EnergySigns for Potential Energy

+6 C

+Q

A

8 cm

B

C

12 cm

4 cm

Consider Points A, B, and C.Consider Points A, B, and C.

For +2 nC at A: For +2 nC at A: U = +1.35 mJU = +1.35 mJ

If +2 nC moves from A to B, does field E do + or – work? Does P.E. increase or decrease?

Questions:

+2 nCMoving

positive q

The field E does positive work, the P.E. decreases.The field E does positive work, the P.E. decreases.

If +2 nC moves from A to C (closer to +Q), the field E does negative work and P.E. increases.

Example 2.Example 2. What is the change in What is the change in potential energy if a potential energy if a +2 nC+2 nC charge charge moves from moves from to to BB??

kQqU

rPotential Potential

Energy:Energy:

2

2

9 -6 -9NmC

(9 x 19 )( 6 x 10 C)(+2 x 10 C)0.900 mJ

(0.12 m)BU

U = -0.450 mJU = -0.450 mJ

Note that P.E. has decreased as work is done by E.

+6 C

+Q

A

8 cm

B

12 cm

From Ex-1: UA = + 1.35 mJ

U = UB – UA = 0.9 mJ – 1.35 mJ

Moving a Negative ChargeMoving a Negative ChargeConsider Points A, B, and C.Consider Points A, B, and C.

Suppose a negative -q is moved.Suppose a negative -q is moved.

If -q moves from A to B, does field E do + or – work? Does P.E. increase or decrease?

Questions:+6 C

+Q

A

8 cm

B

C

12 cm

4 cm

The field E does negative work, the P.E. increases.The field E does negative work, the P.E. increases.

What happens if we move a –2 nC charge from A to B instead of a +2 nC charge. This example follows . . .

Moving negative q -

Example 3.Example 3. What is the change in What is the change in potential energy if a potential energy if a -2 nC-2 nC charge moves charge moves from from to to BB??

kQqU

rPotential Potential

Energy:Energy:

2

2

9 -6 -9NmC

(9 x 19 )(6 x 10 C)(-2 x 10 C)0.900 mJ

(0.12 m)BU

+6 C

+Q

A

8 cm

B

12 cm

From Ex-1: UA = -1.35 mJ

(Negative due to – (Negative due to – charge)charge)

UB – UA = -0.9 mJ – (-1.35 mJ)

U = +0.450 mJU = +0.450 mJ

A – charge moved away from a + charge gains P.E.A – charge moved away from a + charge gains P.E.

Properties of SpaceProperties of Space

E

Electric Field

++++ +

+++Q

.

r

An electric field is a property of space allowing prediction of the force on a charge at that point.

; F

E F qEq

; F

E F qEq

The field E exist independently of the charge q and is found from:

2 :

kQElectric Field E

r 2

: kQ

Electric Field Er

E is a Vector

Electric PotentialElectric Potential

Potential

++++ +

+++Q

.r

Electric potentialElectric potential is another is another property of space allowing us to property of space allowing us to predict the P.E. of predict the P.E. of anyany charge q charge q at a point.at a point.

UV

q

; U

V U qVq

; U

V U qVq

Electric Electric PotentialPotential::

The units are: joules per coulombjoules per coulomb (J/C)(J/C)

For example, if the potential is For example, if the potential is 400 J/C400 J/C at point at point PP, , a a –2 nC–2 nC charge at that point would have P.E. : charge at that point would have P.E. :

U = qV = (-2 x 10-9C)(400 J/C); U = -800 nJU = -800 nJ

P

The SI Unit of Potential The SI Unit of Potential (Volt)(Volt)

From the definition of electric potential as From the definition of electric potential as P.E. per unit chargeP.E. per unit charge, we see that the unit , we see that the unit must be must be J/CJ/C. We redefine this unit as the . We redefine this unit as the volt (V).volt (V).

1 joule; 1 volt =

1 coulomb

UV

q

1 joule; 1 volt =

1 coulomb

UV

q

A potential of one volt at a given point means that a charge of one coulomb placed at that point will experience a potential energy of one joule.

A potential of one volt at a given point means that a charge of one coulomb placed at that point will experience a potential energy of one joule.

Calculating Electric Calculating Electric PotentialPotential

Potential

++++ +

+++Q

.r

kQV

rP

Electric Potential Energy and Electric Potential Energy and Potential:Potential:

; kQq U

U Vr q

; kQq U

U Vr q

kQqr kQ

Vq r

Substituting for Substituting for U, we find V:U, we find V:

kQV

r

kQV

r

The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

Example 4:Example 4: Find the potential at a Find the potential at a distance of 6 cm from a –5 nC charge.distance of 6 cm from a –5 nC charge.

Q = -5 nC

---- -

---Q

.r

P6 cm

2

29 -9Nm

C9 x 10 ( 5 x 10 C)

(0.06 m)

kQV

r

VP = -750 VVP = -750 VNegative V at

Point P :

What would be the P.E. of a –4 What would be the P.E. of a –4 C C charge placed at this point P?charge placed at this point P?

U = qVU = qV = (-4 x 10 = (-4 x 10-6-6 C)(-750 V);C)(-750 V); U = 3.00 mJU = 3.00 mJ

Since P.E. is positive, E will do + work if q is released.Since P.E. is positive, E will do + work if q is released.

q = –4 C

Potential For Multiple Potential For Multiple ChargesCharges

The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

+

- Q1

Q2Q3

-

Ar1

r3

r2

31 2

1 2 3A

kQkQ kQV

r r r

kQV

r

kQV

r

Potential is + or – based on sign of the charges Q.Potential is + or – based on sign of the charges Q.

Example 5:Example 5: Two charges Two charges QQ11= +3 nC= +3 nC and and QQ22 = -5 nC = -5 nC are separated by are separated by 8 cm8 cm. . Calculate the electric potential at point Calculate the electric potential at point AA..

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B1 2

1 2A

kQ kQV

r r

2

29 -9Nm

C1

1

9 x 10 ( 3 x 10 C)450 V

(0.06 m)

kQ

r

2

29 -9Nm

C2

2

9 x 10 ( 5 x 10 C)2250 V

(0.02 m)

kQ

r

VA = 450 V – 2250 V; VA = -1800 VVA = -1800 V

Example 5 (Cont.):Example 5 (Cont.): Calculate the Calculate the electric electric potentialpotential at point at point BB for same charges. for same charges.

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B1 2

1 2B

kQ kQV

r r

2

29 -9Nm

C1

1

9 x 10 ( 3 x 10 C)1350 V

(0.02 m)

kQ

r

2

29 -9Nm

C2

2

9 x 10 ( 5 x 10 C)450 V

(0.10 m)

kQ

r

VB = 1350 V – 450 V; VB = +900 VVB = +900 V

Example 5 (Cont.):Example 5 (Cont.): Discuss meaning of the Discuss meaning of the potentials just found for points potentials just found for points AA and and BB..

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

BVA = -1800 VVA = -1800 V

For every coulomb of positive charge placed at point A, the potential energy will be –1800 J. (Negative P.E.)

For every coulomb of positive charge placed at point A, the potential energy will be –1800 J. (Negative P.E.)

The field holds on to this positive charge. An external force must do +1800 J of work to remove each coulomb of + charge to infinity.

The field holds on to this positive charge. An external force must do +1800 J of work to remove each coulomb of + charge to infinity.

Consider Point Consider Point A:A:

Example 5 (Cont.):Example 5 (Cont.): Discuss meaning of the Discuss meaning of the potentials just found for points potentials just found for points AA and and BB..

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

BVB = +900 VVB = +900 V

For every coulomb of positive charge placed at point B, the potential energy will be +900 J. (Positive P.E.)

For every coulomb of positive charge placed at point B, the potential energy will be +900 J. (Positive P.E.)

Consider Point Consider Point B:B:

For every coulomb of positive charge, the field E will do 900 J of positive work in removing it to infinity.

For every coulomb of positive charge, the field E will do 900 J of positive work in removing it to infinity.

Potential DifferencePotential Difference

The potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

The potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

Potential Difference: VAB = VA - VBPotential Difference: VAB = VA - VB

WorkAB = q(VA – VB) Work BY E-fieldWorkAB = q(VA – VB) Work BY E-field

The positive and negative signs of the charges may be used mathematically to give appropriate signs.

The positive and negative signs of the charges may be used mathematically to give appropriate signs.

Example 6:Example 6: What is the potential difference What is the potential difference between points between points AA and and BB. What work is done by . What work is done by the E-field if a +2 the E-field if a +2 C charge is moved from A C charge is moved from A to B?to B?

VB = +900 VVB = +900 VVA = -1800 VVA = -1800 V

VVABAB= V= VAA – V – VBB = -1800 V – 900 V = -1800 V – 900 V

VAB = -2700 VVAB = -2700 VNote point B is atNote point B is at higher potential.higher potential.

WorkWorkABAB = = q(Vq(VAA – V – VBB) = ) = (2 x 10(2 x 10-6-6 C )(-2700 V) C )(-2700 V)

Work = -5.40 mJWork = -5.40 mJ

Thus, an external force was required to move the charge.

+

-5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B

Q2

E-field does negative work.E-field does negative work.

Example 6 (Cont.):Example 6 (Cont.): Now suppose the +2 Now suppose the +2 C C charge is moved from back from charge is moved from back from BB to to AA??

VB = +900 VVB = +900 VVA = -1800 VVA = -1800 V

VVBABA= V= VBB – V – VAA = 900 V – (-1800 V) = 900 V – (-1800 V)

VBA = +2700 VVBA = +2700 VThis path is from This path is from high to low potential.high to low potential.

WorkWorkBABA = = q(Vq(VBB – V – VAA) = ) = (2 x 10(2 x 10-6-6 C )(+2700 V) C )(+2700 V)

Work = +5.40 mJWork = +5.40 mJ

The work is done BY the E-field this time !

+

-5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B

Q2

E-field does positive work.E-field does positive work.

Parallel PlatesParallel Plates

VA + + + +

- - - -VB

E+q

F = qE

Consider Two parallel plates of equal Consider Two parallel plates of equal and opposite charge, a distance and opposite charge, a distance dd apart.apart.

Constant E field: F = qEConstant E field: F = qE

Work = Work = FdFd = (qE)d= (qE)d

Also, Work = Also, Work = q(Vq(VAA – V – VBB))

So that: So that: qV qVAB AB = qEd = qEd andand VAB = EdVAB = Ed

The potential difference between two oppositely charged parallel plates is the product of E and d.

The potential difference between two oppositely charged parallel plates is the product of E and d.

Example 7:Example 7: The potential difference The potential difference between two parallel plates is between two parallel plates is 800 V800 V. If . If their separ- ation is their separ- ation is 3 mm3 mm, what is the , what is the field field EE??

VA + + + +

- - - -VB

E+q

F = qE

; V

V Ed Ed

; V

V Ed Ed

80 V26,700 V/m

0.003 mE

The E-field expressed in volts per meter (V/m) is known as the potential gradient and is equivalent to the N/C. The volt per meter is the better unit for current electricity, the N/C is better electrostatics.

The E-field expressed in volts per meter (V/m) is known as the potential gradient and is equivalent to the N/C. The volt per meter is the better unit for current electricity, the N/C is better electrostatics.

Summary of FormulasSummary of Formulas

; kQq U

U Vr q

; kQq U

U Vr q

kQV

r

kQV

r

WorkAB = q(VA – VB) Work BY E-fieldWorkAB = q(VA – VB) Work BY E-field

; V

V Ed Ed

; V

V Ed Ed

Electric Potential Energy and Potential

Electric Potential Energy and Potential

Electric Potential Near Multiple charges:

Electric Potential Near Multiple charges:

Oppositely Charged Parallel Plates:

Oppositely Charged Parallel Plates:

CONCLUSION: Chapter 25CONCLUSION: Chapter 25Electric PotentialElectric Potential