Transcript Field Definition And Coulomb`s Law

Field Definition And
Coulomb’s Law
Coulomb’s Law
• Gives us the rule for
dealing with two point
charges.
(in practice for two charges
whose separation is
much greater than the
radius of the charges.)
Q1Q2
F 2
r
r
Charge
Q1
Charge
Q2
The field strength of an electric field
is defined by
F
E
Q
(Unit NC-1)
That is the force exerted by the field on unit charge
(ie a charge of 1 Coulomb)
placed at that point.
+1C
F
With a Radial Field
The electric field strength take
sthe form of Coulombs law. Why?
1
Q
E
40 r 2
+
Q
The two variable quantities are
the charge Q and the distance r
The electric field strength formula is just coulombs law
applied to a test charge of 1C !!
Test
Charge
Uniform electric fields
Test charge
+ +
+
_
_ _
In a uniform field the field strength at any point is given by
Remember this is just the force on a
unit charge in the field
V
E
d
Remember this
only applies
where the field
lines are parallel
Together with these relationships:
W
V
Q
Q
E
40 r 2
The definition of the volt
1
V
E
d
The electric field strength due to a point charge
The electric field strength in a uniform field
So the unit of E is Vm-1 as well as NC-1
Form the basis of any solution to the electric fields questions you will be
asked
Electric Potential
• The electrical potential of any point in the field is the
work done to bring a (+) charge of 1 coulomb from
infinity (i.e. beyond the influence of the field) to that point
in the field.
So the electric potential
at point P
P
Q
1 coulomb
positive
charge
Implications:
1.The electrical potential of any point beyond
the field is zero
2. The electric potential is
the potential energy change for 1C of charge
The electric potentila is given by:
1 Q
V
40 r
Calculations
4.0μC
-6.0μC
A
B
Two charges with the values shown are placed along are separated
100mm from A along the line AB
by a distance of 100mm. At what distance
does the electric potential reach 0V?
When the potential along AB reaches zero
VA=VB
i.e.
Now:
VA +VB = 0
1 QA
VA 
40 rA
1 QB
VB 
40 rB
1 QA
1 QB
0

40 rA 40 rB
1 QA
1 QB

40 rA
40 rB
QA
QB

rA
rB
4.0μC
V=0
40mm
60mm
A
rB
QB

rA
QA
B
100mm
6
rB
 6.0 10
6


6
rA
4.0 10
4
-6.0μC
This ratio tells us that V=0 40mm from A
Where the numbers are not as straightforward you can continue as
follows:
1
2
rA  rB  100
As the total distance between the charges is 100mm
4
rA  rB
6
2
rA  rB
3
2
Now substituting 2 into 1
rB  rB  100
3
5
rB  100
3
300
rB 
 60
5
rA  100  60  40mm
Calculate the magnitude of the electric field strength at the surface of a
nucleus U (Z=92 M=238) . Assume that the radius of this nucleus is 7.4 × 10–15 m.
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Magnitude of electric field strength =.........................................
State the direction of this electric field.
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State one similarity and one difference between the electric field and the
gravitational field produced by the nucleus.
Similarity ....................................................................................................................
. ...................................................................................................................................
Difference ...................................................................................................................
.
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The diagram shows a positively charged oil drop held at rest
between two parallel conducting plates A and B.
A
Oil drop
2.50 cm
B
The oil drop has a mass 9.79 x 10–15 kg. The potential difference between the plates
is 5000 V and plate B is at a potential of 0 V. Is plate A positive or negative?
………………………………………………………………………………………………
Calculate the electric field strength between the plates.
Electric field strength =…………………………………
Calculate the magnitude of the charge Q on the oil drop.
Charge =……………………………………
How many electrons would have to be removed from a neutral oil drop for it to acquire this
charge?
………………………………………………………………………………………………
(3)