Download Electric Charge

Coulomb’s Law
Electricity Equations are Similar to Gravity
m1
Fg = m2 g1
q1
m2
Fg = G
m2 g = G
g1 = G
m1m2
r
m1
r
2
2
m1m2
r
2
q2
FE = q2 E1
q2 E = k
E1 = k
FE = k
q1q2
r
q1
r
2
2
q1q2
r
2
Solving for the Field
m1
P
mnew
q1
qP
new
m
g =G 2
r
q
E=k 2
r
Fg = mnew g
FE = qnew E
Often problems begin by finding the field strength at point P .
Then a new mass/charge appears at point P .
Typically the problem then asks for the force on the new
mass/charge caused by the field of the original mass/charge.
Fields Near Point Masses and Point Charges
A point mass/charge is an object where the field acts as though it is
coming from a single point at the center of the object.
Spheres
M
+
−
Note: The fields around the charges only look like this when they
are very very very far away from any other charge.
We will soon see that the field of a proton and electron next to each
other looks different.
Example 2
A spherical conductor has a charge of 4 μC .
a.
How are the charges distributed
on the conductor?
Excess charge piles up on the
surface of conductors.
b. Determine the magnitude of the electric
field at a point P , located 30 cm to the
right of the charge.
(
q
E = k 2 = 9 ´ 109
r
()
4 ´ 10-6
(0.30)
2
4 μC
P
E
) = 4.0 ´ 10 N C
5
c. State the direction of the electric field.
1st Method:
Charge is positive, and positive means away. field is +x
2nd Method:
Imagine a positive test charge at point P. Now ask which
direction the test charge would move. It will be repelled by the positive
charge, so it will move right, or +x .
Example 2
A spherical conductor has a charge of 4 μC .
A −0.5 μC charge is inserted at point P .
P
E
−0.5 μC
4 μC
FE
d. Determine the magnitude of force
on the point charge
(
)(
)
FE = qE = 0.5 ´ 10-6 4.0 ´ 105 = 0.20 N
e. Determine the direction of force on the point charge.
1st Method:
The new charge is negative, and negative charges move
opposite the electric field ( − means opposite or 180o). The field is +x , so the
force has to be −x .
2nd Method:
If an object is released from rest it will accelerate in the
direction of force. Opposite charges atttract. When the new charge is released
it will be attracted to the positive charge and move to the left, −x .
Coulomb’s Law and the Force of Electricity
Electricity is a field force similar to gravity. The equations follow
the same patterns and they are worth comparing in order to
reinforce each other.
Newton’s Law of
Universal Gravitation
Coulomb’s Law
m1m2
Fg = -G 2
r
q1q2
FE = k 2
r
G = 6.67 ´10
-11
k = 9 ´10
9
Coulomb’s Law and the Force of Electricity
q1q2
FE = k 2
r
k = 9 ´109
This version of Coulomb’s Law and the constant k are used in
most situations in this course.
The formal variant of the formula has certain advantages.
1 q1q2
FE =
4p Î0 r 2
Î0 = 8.85 ´ 10
Where the constant is the permittivity of free space
1
1
9
k=
=
= 9 ´ 10
-12
4p Î0 4p 8.85 ´ 10
(
)
-12
Example 1
Fg = -G
(
Determine the force of gravity and the force of electricity
between a +1.0 C charge with a mass of 1.0 × 10−8 kg and −1.0
C charge with a mass of 5.7 × 10−12 kg separated by 1 m .
m1m2
r
FE = k
2
Fg = 6.67 ´ 10
() 1.0 ´ 10 ) (5.7 ´ 10 )
-8
-11
Fg = 3.8 ´ 10-30 N
(1)
-12
2
FE
q1q2
r2
1.0 ) (1.0 )
(
= ( 9 ´ 10 )
(1)
9
2
FE = 9.0 ´ 109 N
Electricity is many time stronger than gravity.
As a result, gravity is often negligible compared to electricity.
As a rule if you cannot see the charged object with you then its
mass is too small to be significant when compared with gravity.
Newton’s 3rd Law Revisited
FE = k
q1q2
r
2
Is the force calculated by the equations the force of the proton
pulling on the electron, or the electron pulling on the proton ?
It is both. Newton’s 3rd Law states that when two objects interact
there is an equal and opposite force between them.
The electron may be many time smaller than the proton, but it
pulls on the proton with equal force.
The signs in the equations
FE = k
q1q2
r
2
Plugging in positive and negative values for charge q is not
necessary. If you do there are two possible outcomes.
A Positive answer means:
Away, Repel
A Negative answer means: Toward, Attract
We already know that opposite charges attract, and that like
charges repel. We can look at the sign of the charge in the
diagram to decide which way charges will move.
We don’t need the sign from the equation to tell us what is
already obvious.
However, calculus and graphing are done very formally and
the minus signs are important.
The signs in the equations
+
−
+m1 + m2
-Fg = -G
r2
−
+
+q1 - q2
-FE = k
r2
Negative = Toward
Negative = Toward
+
+
+
+m1 + m2
-Fg = -G
r2
+
+q1 + q2
+FE = k
r2
Negative = Toward
Positive = Away
−
−
−
+m1 + m2
-Fg = -G
r2
Negative = Toward
−
+ FE = k
-q1 - q2
r2
Positive = Away
Inverse Square Law
m1m2
Fg = -G 2
r
q1q2
FE = k 2
r
Both the Universal Law of Gravitation and Coulomb’s Law have
distance between the objects in the denominator
Inverse
and the distance is squared
Square
If distance of separation changes, then force also changes by a
factor equal to the inverse square of change in distance.
If the separation distance doubles, then force is ¼ of its original
strength.
If the separation distance is cut in half, then force is 4 times
greater than before.
Motion Perpendicular to a Radial Field
The motion is circular
It is the same for gravity and electricity
In electricity the particles must have opposite charge.
m
M
v0
−
+
v0
+
v0
−
In electricity this is a simplistic way of describing the motion of an
electron around a proton in a hydrogen atom.
Motion Perpendicular to a Radial Field
Solving for any circular motion force problem requires the use of
the circular motion sum of forces.
Gravity
Electricity
Fc = Fg
Fc = FE
v
Mm
m =G 2
r
r
v
Qq
m =k 2
r
r
2
GM
v=
r
2
v=
kQq
mr
Example 3
An electron orbits a proton at a radius of 1.2 × 10−10 m.
Determine the speed of the orbiting electron. Assume a circular
orbit.
Fc = FE
v
Qq
m =k 2
r
r
2
v=
kQq
mr
v=
(
)(
)(
9 ´ 109 1.6 ´ 10-19 1.6 ´ 10-19
(9.11 ´ 10 ) (1.2 ´ 10 )
-31
v = 1.45 ´ 10 m s
6
-10
)