chapter 21 electric charge and electric field...
TRANSCRIPT
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CHAPTER 21
ELECTRIC CHARGE AND
ELECTRIC FIELD
TWO BASIC CONCEPTS:
COULOMB’S LAW
ELECTRIC FIELD
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ATOMS – are neutral
Fig 21.4
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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 21.2
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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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
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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Insert fig 21.29
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Example 21.2 (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
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
Therefore
)
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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(towards Q1)
(away from Q2)
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EA2
EA
A EA1
300
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NOW CALCULATE THE FORCE ON A
CHARGE Q3 = 50µC PLACED AT POINT A.
WHAT DIRECTION?
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HAVE BEEN TALKING ABOUT POINT
CHARGES.
WHAT ABOUT CONTINUOUS CHARGES?
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Consider a ring of uniform charge
Fig 21.24
BY SYMETRY Ey = 0
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Where
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Put in for
Finally
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THIS IS THE ELECTRIC FIELD AS A
FUNCTION OF x FOR A RING OF CHARGE
WITH RADIUS a.
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START WITH THAT EQUATION AND FIND
THE ELECTRIC FIELD ALONG THE AXIS OF
A DISK WITH TOTAL CHARGE Q AND
RADIUS OF R.
INSERT FIG 21.26
ONCE AGAIN BY SYMETRY Ey = 0
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Consider a small ring in the disk as
shown.
For the ring:
Radius = r
Thickness = dr
Charge on ring = dQ
WE CAN WRITE FOR THE CONTRIBUTION
TO THE ELECTRIC FIELD AT P IN THE x
DIRECTION
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TO FIND THE TOTAL FIELD INTEGRATE
USE THE CHARGE DENSITY σ
THEN
CHARGE ON RING
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NOW CONSIDER AN INFINITE SHEET
WITH CHARGE DENSITY σ
WHAT IS THE ELECTRIC FIELD A
DISTANCE x FROM THE SHEET?
FOR A DISK OF RADIUS R
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LET RADIUS EXPAND TO A VALUE OF
INFINITY
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ANYWHERE NEAR AN INFINITE SHEET OF
CHARGE THE ELECTRIC FIELD WILL BE