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Electric Potential
a. Work- Energy Theorem and Potential Energy – review - Recall the
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definition of work from your previous physics courses: W = ∫ F ⋅ ds , as
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the force done by a force F in moving a particle from A to B along a
curve traced by s. Note that this is a line integral that follows a
particular trajectory of the particle. In a closed system where energy
is conserved, we have that K + U = constant, where K and U are the
kinetic and potential energies of the system. This implies that ∆K +
∆U = 0 and therefore that ∆U = - ∆K. This holds in general when all
the forces are conservative, allowing the introduction of potential
energies. But you have previously shown that the work done is equal
to the change in kinetic energy, or W = ∆K, from which we can
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conclude that ∆U = −∆K = − ∫ F ⋅ ds . Recall that the units of work and
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energy are Joules (J).
b. Potential energy of a charge in an electric field - If we consider the
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work done on a charge qo by an electric field, the force is F = qo E so
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the potential energy is given by ∆U = −qo ∫ E ⋅ ds . {At this point it is
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useful to make an analogy with gravitation: with F = mg , the potential
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energy was found from a similar calculation to be ∆U = −m ∫ g ⋅ ds . But
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in this case since g is constant (near the earth’s surface) and always
points vertically downward, the integral is simply equal to -g∆h (with
∆h positive for an upward displacement), because the dot product
ensures that only the vertical displacement contributes to the integral
(with the – sign appearing since g is downward and our displacement
is upward), and our result is that ∆U = mg∆h, a familiar result. In
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more general situations when g is not constant, we would need to use
the general expression from Newton’s law of gravitation that
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g = 2 rˆ .} In the case of electric fields we will see below for
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different situations how to calculate the potential energy.
c. Electric potential. Next, we introduce the electric potential difference,
V, as the potential energy per unit (positive) charge. We can write
that V = U/qo, where only differences in potential are meaningful. In
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other words the electric potential difference between two points can
be written as V(B) – V(A) = ∆V = {U(B) – U(A)}/qo = ∆U/qo. Using
the expression for ∆U in section b we can write that
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∆V = − ∫ E ⋅ ds .
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Note several things about this expression:
• Physically the potential difference is the negative of the work
done by the electric field on the charge qo per unit charge; as
such it is equal to the work done per unit charge by an external
force in order to move the charge qo from point A to point B.
• Potential differences do not depend on the charge qo that is in
the electric field. In other words we can imagine a positive unit
charge that we move around in the electric field to measure the
potential at different points by finding the work done to move
this unit charge about between points, but the potential
differences exist independently of the charge qo.
• Unlike problems with gravity where we often shifted the
arbitrary zero level of potential energy to occur at different
heights, here we will establish the usual convention that the
zero level of electric potential energy is at a point infinitely far
from the charges producing the electric fields. In this case we
can refer to the electric potential and potential energy at a point
V(P) and U(P), where we are always speaking of the zero
reference point as being infinitely far away. We will see the
reason for this choice shortly.
• With this choice for a zero level we can write that
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V ( P) = − ∫ E ⋅ ds , so that V(P) represents the negative of the work
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∞
done by the electric field in bringing the positive unit charge qo
from infinity to point P. This is equal to the positive work
required by an external force to do the same displacement of the
charge.
• Units for V follow from its definition, V(P) = U(P)/qo, as 1 J/C,
which we call a volt (V) so that 1 V = 1 J/C.
• We note here that the units for E, which we previously used 1
N/C, can also be written from the integral expression for
potential as 1 V/m.
• Lastly, a useful unit for dealing with energy on the atomic scale
is the electron volt, defined as the energy an electron gains (or a
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proton loses) by moving through a potential difference of 1 V;
therefore, 1 eV = e(1 V) = 1.6 x 10-19 J.
d. Example. Uniform Electric Field – Suppose we move a charge qo
from A (x1, y1) to B (x2, y2) along the line shown in a uniform electric
field Eo. What is the change in electric potential energy and the
potential difference?
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y
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A
x
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We calculate ∆U = −qo ∫ E ⋅ ds ; in this expression E = Eoiˆ and we can
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recognize that if both E = constant and the cos θ from the dot product
is a constant, we can take them both out of the integral and we are left
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with ∆V = − E ⋅ ∫ ds = − E ⋅ s = − Eo d . Note that as the charge moves with a
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component of its displacement along the E field direction, its potential
energy goes down. The potential difference between the two points is
given by ∆V = ∆U/qo = -Eod. The answer we got does not depend on
the path traveled between A and B.
Equipotential surfaces Note that if we were to move the charge along the y-axis, no work
would be required. In that case we would be moving the charge along
an equipotential surface – defined as a surface all of whose points are
at the same potential. In this example, the equipotential surfaces are
planes perpendicular to the E field and thus to the x-axis – or planes
parallel to the y-z plane. In general, no work is required to move
around on an equipotential surface. Furthermore, the equipotential
surfaces must always be perpendicular to the E field at all points since
if not, there would be a component of E in the plane and the potential
could not be uniform in the plane. In lab #2 you mapped equipotential
surfaces for a variety of situations.
e. Example – Potential due to a single point charge – derivation as in
text
Note that V is a scalar field.
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f. Finding Electric Field from Potential – Assume that we know V(x, y,
z) everywhere. Then we can show that you can find the electric field
everywhere. Before we show the mathematical connection, let’s
discuss the significance of this. Finding V is, in general, much easier
than finding E because V is a scalar field while E is a vector field and
has three vector components, in general, in 3-dimensions. Knowing
how to find E from V then is a major simplification – and almost all
difficult problems are approached by finding V first and then finding
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E from it. How is this done? Well, we know that ∆V = − ∫ E ⋅ ds , so
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then we have that dV = − E ⋅ ds , since integrating both sides gives an
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expression for ∆V. If we pick ds along the x axis so that ds = dx iˆ ,
− dV
then we can write that dV = -Ex dx or that Ex =
. In general we
dx
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 ∂V
∂V ˆ ∂V ˆ 
j+
k = −∇V , where the del operator
have that E = −  iˆ +
∂y
∂z 
 ∂x
yields the gradient of V. As an example, consider the potential of a
ke Q
. Using the gradient operator in
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dV keQ
spherical coordinates, Er = −
= 2 , the expected answer for the
dr
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point charge found above: V =
radial electric field.
g. Potential due to a dipole – also finding E from V; also Work to
assemble charges
h. Example - Potential of a ring of charge along the axis – also finding E
from V
i. Example – Potential of a uniformly charged circular disk along the
axis
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