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CHAPTER 17

ELECTRIC CHARGE AND

ELECTRIC FIELD

TWO BASIC CONCEPTS:

COULOMB’S LAW

� = � ������

ELECTRIC FIELD

� = �

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ATOMS – are neutral

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IONS – are charged

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SOME MATERIALS –

e’s move easily – conductors

metals

SOME MATERIALS –

e’s don’t move easily –

insulators

glass

wood

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FORCES ON CHARGED OBJECTS

LIKE CHARGES REPEL

UNLIKE CHARGES ATTRACT

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USE COULOMB’S LAW

� = � � ����

Example

Two point charges q1 = 25 nC

and q2 = -75 nC are separated

by a distance of 3.0 cm. Find

the magnitude and direction of

the electric force that q1 exerts

on q2.

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� = � � ����

= �9�10�� �+25�10����−75�10���

�0.030���= 0.01875!

Attractive Force

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Also

Newton’s third law – if body A

exerts a force on body B, then

body B exerts an equal and

opposite force on body A.

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ELECTRIC FIELD

Electric force per unit charge.

� = �

Or

� = �"#$"

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In direction that positive charge

would move.

For more than one charge the

total electric field equals the

vector sum of all electric fields

due to each charge.

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For two charges q1 and qtest

� = � % %&'(&��

So

� = �"#$" =

)*�*"#$"+��"#$"

Therefore

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� = � ����

Or

� = � ���

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ALSO

� = ,-./

Where

01 = 8.854�10� � 3�!��4

Therefore can write

� = ,-./

������

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And

� = ,-./

���

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ELECTRIC FIELD LINES

1. IN DIRECTION OF FIELD

(POSITIVE TEST CHARGE

MOVES ALONG LINE.)

2. NUMBER OF LINES

PROPORTIONAL TO

ELECTRIC FIELD.

3. ELECTRIC FIELD LINES START

ON POSITIVE CHARGE AND

END ON NEGATIVE

CHARGE.

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Example (continued)

Find the electric field 3.0 cm

from an electric charge

q1 = +25 nC.

r = 3 cm

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What is E at distance r due to q1

� = ,-./

��� =

,-./�56 1781.19�

� = 2.5�105! 34

DIRECTION ?

Away from positive q1

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If we place a charge -75nC at a

point 3 cm away from q1 what

is the force on this charge?

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The electric field at q2 is

2.5�105! 34

Therefore

� = %�

= �−75�10��3��2.5�105! 34 )

= 0.01875!

WE OBTAINED FORCE IN TWO

WAYS

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Example- Field from two

charges

Q1 = -50µC at x=0.52m, y=0

Q2 = 50µC at x=0, y=0 and

find E at point A (x=0, y=0.3m).

y

. A

300

. . X

Q2 Q1

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E; =k�−50x10�>C�

�0.6m��

= −1.25x10> N C4

(towards Q1)

�C� = DE51F 17GHI�1.9J��

= 5�10>! 34

(away from Q2)

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EA2

EA

∅

A EA1

300

�C6 = �C LMN301= 1.1�10>! 34

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�CO = �C� − �C NPQ301= 4.4�10>! 34

�C = R�C6� + �CO�= S�1.1�10>�� + �4.4�10>��

�C = 4.5�10>! 34

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TUQ∅ = �CO�C6 =

4.4�10>! 341.1�10>! 34

= 4

∅ = 761

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NOW CALCULATE THE FORCE ON A

CHARGE Q3 = 50µC PLACED AT POINT A.

� = %� = �50V10�>3�4.5�10>! 34 = 225!

WHAT DIRECTION?

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ANYWHERE NEAR AN INFINITE SHEET OF

CHARGE THE ELECTRIC FIELD WILL BE

W2X1

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GAUSS’S LAW

TWO BASIC CONCEPTS

ELECTRIC FLUX

AND

GAUSS’S LAW

ELECTRIC FLUX

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Consider a number of guns at the left

shooting to the right.

Bullets

Sheet

Flux of bullets is number of bullets times

area of sheet perpendicular to bullets.

�YZV[\]]^_` = ![\]]^_`�a`b^^_

Turn sheet on edge.

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Bullets

Sheet

Flux now near zero since area of sheet is

parallel to bullets.

Turn sheet to angle.

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Bullets

Sheet

Flux somewhere between previous two

values.

Assign area vector to sheet.

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Bullets

φ

Sheet A

Flux will be

Φ[dee'&( = ![dee'&(aLMNf

OR THINK ABOUT FLUX THE WAY YOUR

BOOK DOES.

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FOR ELECTRIC FLUX

ELECTRIC

FIELD φ

Sheet A

f = �aLMNf

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GAUSS’S LAW

We will state the law then work some

examples. After that we will do an example

to justify the law.

Statement of Gauss’s law:

The flux through a closed surface is equal to

the net charge enclosed by the surface

divided by ε0.

Equation:

g�ah = 4i��'jke

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g�aLMNf = 4i��'jke

� = 14iX1

Therefore

g�aLMNf = 4i 14iX1 �'jke = �'jke

X1

Consider a conducting sphere of radius R

with charge q on the surface.

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Draw a Gaussian surface around this sphere

with radius r.

Apply Gauss’s Law

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g�aLMNf =�'jkeX1

�'j = %

��4i��� = %01

� = 14iX1

%��

The same equation we had for a point

charge.

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Example

Positive electric charge Q is distributed

uniformly throughout the volume of an

insulating sphere with radius R.

a. Find the electric field at point p where

r < R.

b. Find the electric field at point p where

r > R.

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Part a. Radius of Gaussian Surface = r

Gauss’s Law

g�aLMNf =�'jkeX1

Need charge enclosed by surface.

Charge density l = _m&neopn�q'_m&nermedJ' = �

st-ut

Charge enclosed by Gaussian Surface

�'j = l�vMwx�yyQLwMNyz

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�'j = l {43i�9| =�

43i}9{43i�9| =

�9}9 �

g�aLMNf =�9}9 �01

��4i��� =�9}9 �01

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� =�9}9 �01

4i��

� = 14i01

��}9

For r < R

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Part b.

Draw Gaussian Surface outside sphere.

g�aLMNf =�'jkeX1

�'j = �

Same as before

��4i��� = �01

� = 14i01

���

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