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Voltage ( = Electric Potential )
An electric charge alters the space around it. Throughout the space around every charge is a
vector thing called the electric field. Also filling the space around every charge is a scalar thing,
called the voltage or the electric potential. Electric fields and voltages are two different ways to
describe the same thing.
(Note on terminology: The text book uses the term "electric potential", but it is easy to confuse
electric potential with "potential energy", which is something different. So I will use the term
The voltage at a point in empty space is a number (not a vector) measured in units called volts
(V) . Near a positive charge, the voltage is high. Far from a positive charge, the voltage is low.
Voltage is a kind of "electrical height". Voltage is to charge like height is to mass. It takes a lot
of energy to place a mass at a great height. Likewise, it takes a lot of energy to place a positive
charge at a place where there voltage is high.
lower voltage here
higher voltage here
Only changes in voltage ∆V between two different locations have physical significance. The
zero of voltage is arbitrary, in the same way that the zero of height is arbitrary.
We define ∆V in 2 equivalent ways:
• ∆=V ∆U of q= change in potential energy of a test charge divided by the test charge
B K K
• ∆V=V B A = − E⋅∫r
For constant E-field, this integral simplif∆Vs =− E⋅∆ r (∆r = change in position)
The electric field is related to the voltage in this way: Electric field is the rate of change of
voltage with position. E-field is measured in units of N/C, which turn out to be the same as
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volts per meter (V/m). E-fields points from high voltage to low voltage. Where there is a big E-
field, the voltage is varying rapidly with distance.
In order to understand these strange, abstract definitions of voltage, we must review work and
Work and Potential Energy (U)
Definition of work done by a force: consider an object pulled or pushed by a constant force .
While the force is applied, the object moves through a displacement of ∆r = f- ri
( f ) Notice that the direction of displacement is not the
( i ) ∆r same as the direction of the force, in general.
Work done by a force F = WFF≡⋅ ∆r = F∆rcosθ = F || (constant F)
F = component of force along the direction of displacement
If the force F varies during the displacement (or the displacement is not a straight line), then we
W F ≡ ⋅dr
must use the more general definition of work done by a force F ∫
Work is not a vector, but it does have a sign (+) or (-). Work is positive, negative, or zero,
depending on the angle between the force and the displacement.
θ < 90, W positive θ = 90, W = 0 θ > 90, W negative
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Definition of Potential Energy U: Associated with conservative forces, such as gravity and
electrostatic force, there is a kind of energy of position called potential energy. The change in
potential energy ∆U of a system is defined to be the negative of the work done by the "field
force", which is the work done by an "external agent" opposing the field.
∆U ≡ W ext= −W field
This is best understood with an example: A book of mass
vf= 0 Forces on book:
m is lifted upward a height h by an "external agent" (a hand
which exerts a force to oppose the force of gravity). The g
force of gravity is the "field". In this case, the work done m
by the hand is W ext= +mgh. The work done by the field vi= 0
(gravity) is Wfield −mgh. The change in the potential
energy of the earth/book system is ∆U = W ext= −W field +mgh . The work done by the
external agent went into the increased gravitational potential energy of the book. (The initial and
final velocities are zero, so there was no increase in kinetic energy.)
A conservative force is force for which the amount of work done depends only on the initial and
final positions, not on the path taken in between. Only in the case that the work done by the field
in independent of the path, does it make any sense to associate a change in energy with a change
Potential energy is a useful concept because (if there is no friction, no dissipation)
∆K + ∆U = 0 ⇔ K + U = constant (no dissipation)
(K = kinetic energy = ½ m v 2)
We define electrostatic potential energy (not to be confused with electrostatic potential or
voltage) in the same way as we defined gravitational potential energy, with the relation ∆U =
W ext= −W fieldConsider two parallel metal plates (a capacitor) with equal and opposite charges
on the plates which create a uniform electric field between the plates. The field will push a test
charge +q toward the negative plate with a constant force of magnitude F = q E. (The situation is
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much like a mass in a gravitational field, but there is no gravity in this example.) Now imagine
grabbing the charge with tweezers (an external agent) and pulling the charge +q a displacement
∆r against the electric field toward the positive plate. By definition, the change in electrostatic
potential energy of the charge is
K K K K
∆UU = f−U =i W ext −W field −F⋅fieldr = −q ⋅ E ∆r
I recommend that you do not try to get the signs from the equations – it's too easy to get
confused. Get the sign of ∆U by asking whether the work done by the external agent is positive
or negative and apply ∆U = +W .
( f ) hi PE
E F = q E ∆r
( i )
If the E-field is not constant, then the work done involves an integral
f Kf K
∆UU = f i = −W field− ⋅ F fieldr = −q ⋅ E dr .
Now we are ready for the definition of voltage difference between two points in space. Notice
that the change in PE of the test charge q is proportional to q, so the ratio ∆U/q is independent of
G F on q
q. Recall that electric field is defined as the force per charge : . Similarly, we define