Download Lesson 7 (1) Definition of Electric Potential Consider the electric field

Lesson 7
(1) Definition of Electric Potential
Consider the electric field E created by a charge distribution. A test charge q placed
at any location experiences the force
F = qE
where E is evaluated at that location. When
the test charge undergoes a small
displacement dr , the work done by the
electric force is
dW = F × dr = qE × dr
If the test charges moves from the point A to B along a path, the work done can be
calculated by dividing the path into small displacements and adding the work done
along the displacements. In the limit when these displacements go to zero, we find
the work done along the path from A to B in the form of a line integral:
B
W = å dW = q ò E × dr
A
It turns out that the electrostatic force is conservative: the work done in going from
one point to the other is independent of the path between the two points. Further,
since the quantity W q is independent of the test charge, being a property of the
electric field alone, we can define a property of the electric field at any point called
the electric potential so that if VA and VB denote the potentials at A and B, the
difference is
VA -VB =
B
ò E × dr
A
We can only define the potential difference between two points. The absolute value
depends on choosing a reference point where we can assign an arbitrary value to V .
In terms of the potentials, we can write
Work done by electric field = q (VA -VB )
The unit of potential is the Volt (V), which is Joule per Coulomb (J/C).
1
Customarily, in going from A to B, we define the increment of potential as
DV = VB -VA
Therefore
B
DV = - ò E × dr
A
We can imagine applying to a test charge an external force of the same amount as
the electric force but in the opposite direction (push against the electric influence).
Such a force is -qE , and it can be considered to do work against the electric field.
We can write
Work done against electric field = qDV
Thus, in moving a positive charge from a point of low potential to one of high
potential, the external force has to do positive work. However, if the charge is
negative, the external force does negative work.
(2) Uniform Electric Field
Consider a uniform electric field in the x-direction:
E = Exiˆ
and two points A and B on the x-y plane with coordinates ( xA , yA ),
( xB, yB )
respectively. To calculate the work done by the electric field on a test charge of 1C
moving from A to B, divide the path into small displacements dr . We can write
dr = dx iˆ + dy ĵ
ˆ ĵ are
The dot products of the unit vectors i,
iˆ × iˆ = ĵ × ĵ =1
iˆ × ĵ = ĵ × iˆ = 0
The elemental work is
E × dr = Exiˆ × dr = Ex dx
Note that the change in y-coordinate does not enter. The total work is
2
B
ò E × dr =
A
xB
òE
x
dx = Ex ( xB - x A )
xA
which demonstrates the path independence of the line integral. Thus,
B
DV = - ò E × dr = -Ex Dx
A
where Dx = xB - xA . The argument is the same in three dimensions. From the above
relation between electric field and potential difference, we can use volt per meter
(V/m) as the unit of electric field instead of N/C.
Locations where the potentials are the same are
said to form an equipotential surface. The
equipotential surfaces for a uniform electric field
are planes perpendicular to the field lines. The
electric field points from an equipotential surface
of high potential to one of low potential.
(3) Conservation of Energy in the motion of a charged particle in Electric Field
Consider the motion of a charged particle acted on by an electric force alone. Let m
and q be the mass and charge of the particle. According to the work energy
theorem, in moving from the point A to B, the change in kinetic energy
1
1
DK = K B - K A = mu B2 - mu A2
2
2
is equal to the work done by the electric force. Since this work is given by
W = q (VA -VB ) = -qDV
we have K B - K A = q (VA -VB ) , which can be written
KB + qVB = K A + qVA
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This is the statement of conservation of energy if we define the electrostatic energy
of the particle by
U = qV
The conservation of energy can also be expressed in the form
DK + qDV = 0
The kinetic energy of a positively charged particle increases going from high to low
potential, whereas that of a negatively charged particles increases going from low to
high potential. Thus an electric field accelerates a positive charge from high to low
potentials, but does so for a negative charge from low to high potentials.
Instead of using J as the unit of energy, for atomic particles, a unit called the
electron volt (eV) proves to be more convenient.
1 eV =1.6 ´10-19 J = e J
Example: A proton starts at rest at a point where the potential is 200V. What is its
kinetic energy in J and in eV and what is its velocity when it reaches a point where
the potential is -100V?
Solution:
DV = -100 - 200 = -300V
K f = DK = -qDV = -1.6 ´10-19 ´ (-300) = 4.8´10-17 J =
u =
2K f
mp
=
4.8´10-17
eV = 300eV
1.6 ´10-19
2 ´ 4.8´10-17
=2.4 ´10 6 m / s
-27
1.67 ´10
Example: What is the gain in kinetic energy in eV of Al +++ moving between the same
two points as the previous problem? What is the velocity of the ion if it starts from
rest?
Solution:
K f = DK = 3´ 300 = 900eV
 
2K f
mp

2  900  1.6  10 19
 8.0  10 4 m / s
 27
27  1.67  10
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