03 Electric Potential.pptx

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Objectives:

After completing this module,

you should be able to:

• _{Understand and 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

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SIGN CONVENTIONS

Positive Work done by an external force

increases the Potential Energy.

d

F E

Positive Work done by the

field

decreases the Potential Energy.

d

F E

Negative Work done by an external force

decreases the Potential Energy.

d

F E

Negative Work done by the

field

increases the Potential Energy.

d

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Gravitational Work and Energy

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

A B

h mF

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.

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+ + + +

++

Q

+ +

+ + + + + + + + + + + + + +

-Electric Field

weakens with

distance.

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Electrical Work and Energy

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

Work = Fd = (qE)d

At 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

F_e

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Work and Negative Charges

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 d qE

-q

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𝑦

=

1

𝑥

2

𝑦

=

1

∙

4

.25

Find the average height from 1 to 4.

𝑦= 1

1∙1

𝑦= 1

4∙ 4

𝒚=𝟏

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r

_A

r

_B

F_g = G

m₁·m₂

r_B2

F_g = G

m₁·m₂

r_A2

Work = average force · distance

= · (

G

r

_B

–

r

_A

)

m₁·m₂

r_Ar_B

= G·m₁·m₂

1 1

r_Ar_B

= G·m₁·m₂

1 1

∞r_B

zero

= - G·m₁·m₂ rB

Absolute Potential

Energy

U

G

=

G m1·m2

r_B

r_Ar_B

r_A

r_Ar_B

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-2,000 Joules

-8,000 Joules

ΔU_G = Final - Initial

ΔU_G = -8,000 J – (-2,000 J)

ΔU_G = -6,000 Joules

Object LOST 6,000 Joules

ΔU_G = Final - Initial

ΔU_G = -2,000 J – (-8,000 J)

ΔU_G = +6,000 Joules

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Work to Move a Charge

+ ++ + + +++ Q ¥ qE F +

Work to move

+q from A to B.

· · A B r_a r_b At A: At B:

Avg. Force: _Distance:_r_a_{- r}_b

2 a a kqQ F r  avg a b kqQ F r r  2 b b kqQ F r 

( _a _b)

a b kQq

Work Fd r r r r

   1 1

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Absolute Potential Energy

+ ++ + + +++ Q ¥ qE F + · · A B r_a r_b

Absolute P.E. is relative to ¥. It is work to bring

+q from infinity to point near Q—i.e., from ¥ to r_b

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

+6 mC +Q

·

A

+2 nC

Potential Energy:

The P.E. will be positive at point A, because the field can do + work if q is released.

U = 1.35 mJ

U = 1.35 mJ Positive potential energy

8 cm

kQq

U

r



2 2

9 _Nm -6 -9

C

(9 x 10 )( 6 x 10 C)(+2 x 10 C) (0.08 m)

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Signs for Potential Energy

+6 mC +Q

·

A

8 cm

·

B

C

12 cm

4 cm

Consider Points A, B, and C. For +2 nC at A: U = +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 nC Moving

positive q

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

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Example 2. What is the change in potential energy if a

+2 nC charge moves from A to B?

Potential Energy:

DU = -0.450 mJ

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

+6 mC +Q · A 8 cm ·B 12 cm

From Ex-1: U_A = + 1.35 mJ

DU = U_B – U_A = 0.9 mJ – 1.35 mJ

kQq

U

r



2 2

9 _Nm -6 -9

C

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

0.900 mJ (0.12 m)

B

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Moving a Negative Charge

Consider Points A, B, and C.

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 mC +Q

·

A

8 cm

·

B

C

12 cm

4 cm

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

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-Example 3. What is the change in potential energy if a

-2 nC charge moves from A to B?

Potential Energy:

+6 mC +Q · A 8 cm ·B 12 cm

From Ex-1: U_A = -1.35 mJ

(Negative due to – charge)

U_B – U_A = -0.9 mJ – (-1.35 mJ) DDU = +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.

kQq

U

r



2 2

9 _Nm -6 -9

C

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

0.900 mJ (0.12 m)

B

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Properties 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.

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

E= kQq r2 q ; F

E F qE

q

 

2

:

kQ

Electric Field

E

r

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20 J/kg

10 J/kg

60 Joules total

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Electric Potential

Potential +

++ + +

+++

Q

.

r

Electric potential is another property of space allowing us to predict the P.E. of

any charge q at a point.

Electric

Potential:

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

For example, if the potential is 400 J/C at point P, a –2 nC

charge at that point would have P.E. :

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

P V U

q



;

U

V

U

qV

q

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The SI Unit of Potential (Volt)

From the definition of electric potential as P.E. per unit charge, we see that the unit must be

J/C. We redefine this unit as the volt (V).

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.

1 joule

; 1 volt =

1 coulomb

U

V

q











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Calculating Electric Potential

Potential + ++ + + +++ Q . r P Electric Potential Energy and

Potential:

Substituting for U, we find V:

The potential due to a positive charge is positive; The potential due to a negative charge is negative.

(Use sign of charge.)

The potential due to a positive charge is positive; The potential due to a negative charge is negative.

(Use sign of charge.)

kQ V r 

;

kQq

U

V

r

q



 

kQq

r kQ

V

q r

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

Q = -5 nC -- - --Q . r P 6 cm

V_P = -750 V V_P = -750 V

Negative V at Point P :

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

U = qV = (-4 x 10-6 mC)(-750 V); U = 3.00 mJ_U_{= 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 mC





2 2

9 _Nm -9

C

9 x 10 ( 5 x 10 C)

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Potential For Multiple Charges

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

Q₂ Q₃

-A r₁

r₃ r2

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

3 1 2

1 2 3 A

kQ

V

r



+

kQ

V

r

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Example 5: Two charges Q1= +3 nC and Q2 = -5 nC are separated by 8 cm. Calculate the electric potential at point A.

+

·

Q₂= -5 nC

-Q₁ ₊_{3 nC}

· 6 cm 2 cm 2 cm A B

V_A = 450 V – 2250 V; V_V_A_A = -1800 V_{= -1800 V}

1 2 1 2 A

kQ

V

r



+



2



2

9 _Nm -9

C 1

1

9 x 10 ( 3 x 10 C)

450 V (0.06 m) kQ r +   +



2



2

9 _Nm -9

C 2

2

9 x 10 ( 5 x 10 C)

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Example 5 (Cont.): Calculate the electric potential at point B

for same charges.

+

·

Q₂= -5 nC

-Q₁ ₊_{3 nC}

· 6 cm 2 cm 2 cm A B

V_B = 1350 V – 450 V; _V

B = +900 V

V_B = +900 V

1 2 1 2 B

kQ

V

r



+



2



2

9 _Nm -9

C 1

1

9 x 10 ( 3 x 10 C)

1350 V (0.02 m) kQ r +   +



2



2

9 _Nm -9

C 2

2

9 x 10 ( 5 x 10 C)

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

+

·

Q₂= -5 nC

-Q₁ ₊_{3 nC}

· 6 cm 2 cm 2 cm A B V_A = -1800 V

V_A = -1800 V

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.

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

+

·

Q₂= -

-

5 nC

Q₁ ₊_{3 nC}

·

6 cm

2 cm 2 cm

A B V_B = +900 V

V_B = +900 V

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

Consider Point B:

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

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Potential 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.

Potential Difference: V_AB = V_A - V_B

Potential Difference: V_AB = V_A - V_B

Work_AB = q(V_A – V_B) Work BY E-field

Work_AB = q(V_A – V_B) 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

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Example 6: What is the potential difference between points

A and B. What work is done by the E-field if a +2 mC charge is moved from A to B?

V_B = +900 V V_B = +900 V V_A = -1800 V

V_A = -1800 V

V_AB= V_A – V_B = -1800 V – 900 V

V_AB = -2700 V

V_AB = -2700 V Note point B is at_{higher potential.}

Work_AB = q(V_A – V_B) = (2 x 10-6 C )(-2700 V)

Work = -5.40 mJ Work = -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

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Example 6 (Cont.): Now suppose the +2 mC charge is moved back from B to A?

V_B = +900 V V_B = +900 V V_A = -1800 V

V_A = -1800 V

VBA= VB – VA = 900 V – (-1800 V)

V_BA = +2700 V

V_BA = +2700 V This path is from _{high to low potential.}

Work_BA = q(V_B – V_A) = (2 x 10-6 C )(+2700 V)

Work = +5.40 mJ Work = +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

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+ + + +

++

Q

+ +

+ + + + + + + + + + + + + +

-Electric Field

weakens with

distance.

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Parallel Plates

V_A _{+ + + +}

-V_B

E

+q

F = qE

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

Constant E field: F = qE

Constant E field: F = qE

Work = Fd= (qE)d

Also, Work = q(V_A – V_B)

So that: qV_AB= qEd and V_VAB_AB = Ed_{= Ed}

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+ + + +

++

Q

+ +

Electric Field

weakens with distance.

Electric Field UNIFORM

+ + + + + + + + + + + + + +

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-Example 7: The potential difference between two parallel plates is 80 V. If their separation is 3 mm, what is the field E?

V_A + + + +

-V_B

E

+q

F = qE

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.

;

V

Ed

E

d



80 V

26,700 V/m 0.003 m

E  

N ; Units

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E =

F

q

=

V

r

q =

F·r

V

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Equal

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Equipotential

Lines

g - Field

(J/kg)

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Equipotential

Lines

E - Field

(J/C)

(N/C)

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Lake Tahoe

1,645 ft deep

Equipotential

Lines

g - Field

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Lake Tahoe

1,645 ft deep

Equipotential

Lines

g - Field

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http://son.nasa.gov/tass/images/electric_fields3.jpg

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-Summary of Formulas

Work_AB = q(V_A – V_B) Work BY E-field

Work_AB = q(V_A – V_B) Work BY E-field Electric Potential

Energy and Potential Electric Potential Energy and Potential

Electric Potential Near Multiple charges:

Oppositely Charged Parallel Plates:

;

kQq

U

V

r

q



kQ

V

r





;

V

Ed

E

d