CHAPTER 18 ELECTRIC POTENTIAL - Texas A&M Universitypeople.physics.tamu.edu/adair/phys202/CHAPTER 18.pdf · 6 Now a test charge q0 is placed at position a a distance ra from q0.Then

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CHAPTER 18

ELECTRIC POTENTIAL

BASIC CONCEPTS:

ELECTRIC POTENTIAL ENERGY

ELECTRIC POTENTIAL

ELECTRIC POTENTIAL GRADIENT –

POTENTIAL DIFFERENCE

POTENTIAL ENERGY

2

h

PE = U = mgh

PE KE

Or

U K

And U + K = total energy = constant

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BOOK EXAMPLE

4

Charged Particle in Electric Field is similar

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Consider a point charge q that sets up an

electric field in space

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Now a test charge q0 is placed at position a

a distance ra from q0. Then q0 moves to

position b a distance rb from q0.

What is the change in potential energy?

The change in potential energy is the

negative of the work done to move the test

charge from a to b.

The force on the test charge is

� = 14���

���

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The work done is force times distance. But

the force changes as q0 moves away from q

See Figure 18.6

Must integrate (Not necessary to

understand but the procedure is in the box

on the next page.)

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����� = � ��� =��

� 14���

�����

= ���4��� �1� −

1��

Δ� = �� − �� = − ���4��� �1� −

1��

Use �� = ���

Then

The change in potential energy,Δ�, is the

negative of this work.

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So difference in potential energy between

two points is

Δ� = �� − �� = − ���4��� �1� −

1��

Now define the potential energy to be zero

when the two charged particles are

separated by an infinite distance (b = ∞; Ub

= 0).

0 − �� = − ���4��� �1� −

1∞�

�� = ���4���1�

Or

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�� = ���4���1�

Use �

� !" = # and replace the symbol a

with r the distance from the charge

providing the field to the charge.

� = # $$%�

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ELECTRIC POTENTIAL

Start with:

Difference in potential energy is

Δ� = �� − �� = − ���4��� �1� −

1��

DEFINITION:

ELECTRICAL POTENTIAL IS POTENTIAL

ENERGY PER UNIT CHARGE

&'()*+)�',-*(.*+�'= ,-*(.*+�'&.(/0

�.+*1ℎ�/(

Therefore divide all terms by ��

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���� −���� =

− ���4��� 31� −

1�4��

Thus

5� − 5� = − �4��� �

1� −

1��

5� − 5� = �4��� �

1� −

1��

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Then for potential at a point

The potential energy at a point is

� = # ��6

&'()*+)�',-*(.*+�'= ,-*(.*+�'&.(/0

�.+*1ℎ�/(

U= # $$%�

5 = ��6 = # ��6�6 = # �

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POTENTIALS ADD (SCALERS)

Just as we did with the electric field we can

add the potentials for many charges in an

area.

EXAMPLE

A ∙ 60cm

30cm

8� ∙ ∙ 8

50µC -50 µC

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What is the potential at A?

59 = 59: + 59<

59 = 14�=�

[email protected] + 1

4�=�−50?10@A1

0.6D

59 = 1.5?10A5 − 7.5?10G5 = 7.5?10G5

ELECTRON VOLT

An electron volt is a unit for energy. It is

the work necessary to move an electron

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(charge ( = 1.6?10@�H1) a potential

difference of 1 volt.

1 Volt Batt

The work to move a charge � across a

potential difference is � = �5

� = �5 = I1.6?10@�H1JI15J= 1.6?10@�HK

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Therefore

1(5 = 1.6?10@�HK

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CAPACITANCE AND

DIELECTRICS

MAJOR TOPICS

Calculating Capacitance

Capacitors in Circuits

Energy Storage in Capacitors

Dielectrics

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Capacitors store:

Charge

Energy

+Q + - -Q

+ -

+ -

+ -

+ V _

BATTERY

+ v -

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The battery supplies charge to the plates.

The charge Q is proportional to V.

8 ∝ 5

Choose the proportionality constant C.

8 = 15

C is the CAPACITANCE of the capacitor.

1 = 85

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If there is charge on a plate

there is an electric field E.

+Q + - -Q

+ P -

+ -

+ -

+ V _

BATTERY

E at P due to + plate &M = NO"

+ v -

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E at P due to - plate &@ = NO"

Both fields point to the right

d

+Q + - -Q

+ P -

+ E -

+ -

+ V _

& = NO" + N

O" = NO"

+ v -

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For all capacitors 1 = PQ

And for parallel plate capacitors

1 = ��R�

Parallel plate capacitors are easy:

Area and distance between plates gives C.

If know C then:

Know Q can get V

Know V can get Q

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CAPACITORS IN CIRCUITS

Series

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5 = 5� + 5

1 = 85STUVWXYZZZ[ 5 = 8

1

81U$ =

81� +

81

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11U$ =

11� +

11 \-](+(]

Parallel

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8 = 8� + 8

1 = 85STUVWXYZZZ[ 8 = 15

1U$5 = 1�5 + 15

1U$ = 1� + 1\-^��''('

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ENERGY IN CAPACITOR AND ELECTRIC FIELD

The potential difference across the

capacitor is

5 = ��

Where U is the energy stored in the electric

field of the capacitor and q is the charge on

the plates.

Thus the change in potential difference is

Δ5 = Δ��

Now charge a capacitor

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dq

ΔW=VΔq

5 = $_

Δ� = �1 Δ�

Add up all of the work done to charge the

capacitor to charge q

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� = 11� ��� = 1

1 `�2 b�

PP�

� = 11 `82 − 02b

� = 1281

Once again we use calculus but only to show you how something is done. To

“add up” the work.

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� = 1281

This is the amount of work to charge the

capacitor from 0 charge to a charge of Q.

This is the energy stored in the capacitor.

� = 1281

Use definition of capacitance 1 = PQ

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� = �15

Or � = �85

Then for parallel plate capacitors

1 = !"9W and 5 = &�

� = 1215 =

12��R� &�

� = 12 ��&R�

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R� = c-'dD(-\)�^�)+*-

Energy density d = eQ

Therefore d = eQ =

:<!"f<9W9W = �

��&

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DIELECTRICS

Add material between plates and C

increases.

Increases by K the dielectric constant

Parallel Plate Capacitor

1 = =� R� �Ug�hUXYZZZZZ[ i =� R�

Define � = i=�

Then can use 1 = � 9W

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Other quantities

Energy density

d = 12 =�&

�Ug�hUXYZZZZZ[12i=�&=12 �&

Charge on capacitor connected to V.

V 8� = 1�5

Introduce dielectric

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V dielectric

1 = i1�

1 = i P"Q

15 = i8�

15 = 8 = i8�

Insert dielectric and Q increases if V

remains constant.

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Voltage across capacitor without battery.

1� = PQ"

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1 = i1�

1 = P"Q

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i1� = 8�5

5 = P"_"

�j

But P"_" = 5�

So 5 = Q"j

When C not connected to battery inserting

dielectric decreases V.

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Electric Field in dielectric.

&k = QW =

Q" jlW = Q" Wl

j = f"j

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CAPACITOR SUMMARY

11U$ =

11� +

11 \-](+(]

1U$ = 1� + 1\-^��''('

) = 85

ENERGY

� = 1281

� = 1215

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� = 1285

ENERGY DENSITY

d = �5

d = 12 ��&

DIELECTRICS

� = i=�

1 = i1�

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1 = i8�5

5 = 5�i

&k = &�i