Define equipotential surfaces and equipotential lines
Explain the relationship between equipotential lines and
electric field lines
Map equipotential lines for one or two point charges
Describe the potential of a conductor
Compare and contrast equipotential lines and elevation lines on
topographic maps
We can represent electric potentials (voltages) pictorially,
just as we drew pictures to illustrate electric fields. This is not
surprising, since the two concepts are related. Consider Figure
\(\PageIndex{1}\), which shows an isolated positive point charge
and its electric field lines, which radiate out from a positive
charge and terminate on negative charges. We use blue arrows to
represent the magnitude and direction of the electric field, and we
use green lines to represent places where the electric potential is
constant. These are called equipotential
surfaces in three dimensions, or
equipotential lines
in two dimensions. The term equipotential is also used as
a noun, referring to an equipotential line or surface. The
potential for a point charge is the same anywhere on an imaginary
sphere of radius r surrounding the charge. This is true
because the potential for a point charge is given by \(V = kq/r\)
and thus has the same value at any point that is a given distance
r from the charge. An equipotential sphere is a circle in
the two-dimensional view of Figure \(\PageIndex{1}\). Because the
electric field lines point radially away from the charge, they are
perpendicular to the equipotential lines.
Figure \(\PageIndex{1}\): An isolated point charge Q
with its electric field lines in blue and equipotential lines in
green. The potential is the same along each equipotential line,
meaning that no work is required to move a charge anywhere along
one of those lines. Work is needed to move a charge from one
equipotential line to another. Equipotential lines are
perpendicular to electric field lines in every case. For a
three-dimensional version, explore the first media
link.
It is important to note that equipotential lines are always
perpendicular to electric field lines. No work is required to
move a charge along an equipotential, since \(\Delta V = 0\). Thus,
the work is
\[W = - \Delta U = - q\Delta V = 0.\]
Work is zero if the direction of the force is perpendicular to
the displacement. Force is in the same direction as \(E\), so
motion along an equipotential must be perpendicular to \(E\). More
precisely, work is related to the electric field by
Note that in Equation \ref{eq5}, \(E\) and \(F\) symbolize the
magnitudes of the electric field and force, respectively. Neither
\(q\) nor \(E\) is zero and \(d\) is also not zero. So \(\cos \,
\theta\) must be 0, meaning \(\theta\) must be \(90^o\). In other
words, motion along an equipotential is perpendicular to
E.
One of the rules for static electric fields and conductors is
that the electric field must be perpendicular to the surface of any
conductor. This implies that a conductor is an equipotential
surface in static situations. There can be no voltage
difference across the surface of a conductor, or charges will flow.
One of the uses of this fact is that a conductor can be fixed at
what we consider zero volts by connecting it to the earth with a
good conductor—a process called
grounding. Grounding can be a useful
safety tool. For example, grounding the metal case of an electrical
appliance ensures that it is at zero volts relative to Earth.
Figure \(\PageIndex{2}\): The electric field lines and
equipotential lines for two equal but opposite charges. The
equipotential lines can be drawn by making them perpendicular to
the electric field lines, if those are known. Note that the
potential is greatest (most positive) near the positive charge and
least (most negative) near the negative charge. For a
three-dimensional version, explore the first media
link.
Because a conductor is an equipotential, it can replace any
equipotential surface. For example, in Figure \(\PageIndex{2}\), a
charged spherical conductor can replace the point charge, and the
electric field and potential surfaces outside of it will be
unchanged, confirming the contention that a spherical charge
distribution is equivalent to a point charge at its center.
Figure \(\PageIndex{2}\) shows the electric field and
equipotential lines for two equal and opposite charges. Given the
electric field lines, the equipotential lines can be drawn simply
by making them perpendicular to the electric field lines.
Conversely, given the equipotential lines, as in Figure
\(\PageIndex{2a}\), the electric field lines can be drawn by making
them perpendicular to the equipotentials, as in Figure
\(\PageIndex{2b}\).
Figure \(\PageIndex{3}\): (a) These equipotential lines
might be measured with a voltmeter in a laboratory experiment. (b)
The corresponding electric field lines are found by drawing them
perpendicular to the equipotentials. Note that these fields are
consistent with two equal negative charges. For a three-dimensional
version, play with the first media link.
To improve your intuition, we show a three-dimensional variant
of the potential in a system with two opposing charges. Figure
\(\PageIndex{4}\) displays a three-dimensional map of electric
potential, where lines on the map are for equipotential surfaces.
The hill is at the positive charge, and the trough is at the
negative charge. The potential is zero far away from the charges.
Note that the cut off at a particular potential implies that the
charges are on conducting spheres with a finite radius.
Figure
\(\PageIndex{4}\): Electric potential map of two opposite
charges of equal magnitude on conducting spheres. The potential is
negative near the negative charge and positive near the positive
charge. This dynamic image is powered by CalcPlot3D and can be
viewed
here.
A two-dimensional map of the cross-sectional plane that contains
both charges is shown in Figure \(\PageIndex{5}\). The line that is
equidistant from the two opposite charges corresponds to zero
potential, since at the points on the line, the positive potential
from the positive charge cancels the negative potential from the
negative charge. Equipotential lines in the cross-sectional plane
are closed loops, which are not necessarily circles, since at each
point, the net potential is the sum of the potentials from each
charge.
Figure \(\PageIndex{5}\): A cross-section of the
electric potential map of two opposite charges of equal magnitude.
The potential is negative near the negative charge and positive
near the positive charge.
Note
View this simulation to observe and modify the equipotential
surfaces and electric fields for many standard charge
configurations. There’s a lot to explore.
One of the most important cases is that of the familiar parallel
conducting plates shown in Figure \(\PageIndex{6}\). Between the
plates, the equipotentials are evenly spaced and parallel. The same
field could be maintained by placing conducting plates at the
equipotential lines at the potentials shown.
Figure \(\PageIndex{6}\): The electric field and
equipotential lines between two metal plates. Note that the
electric field is perpendicular to the equipotentials and hence
normal to the plates at their surface as well as in the center of
the region between them.
Consider the parallel plates Figure \(\PageIndex{6}\). These
have equipotential lines that are parallel to the plates in the
space between and evenly spaced. An example of this (with sample
values) is given in Figure \(\PageIndex{6}\). We could draw a
similar set of equipotential isolines for
gravity on hills . If the hill has any extent at the same
slope, the isolines along that extent would be parallel to each
other. Furthermore, in regions of constant slope, the isolines
would be evenly spaced. An example of real topographic lines is
shown in Figure \(\PageIndex{7}\).
Figure \(\PageIndex{6}\). (a) A topographical map of
Devil’s Tower, Wyoming. Lines that are close together indicate very
steep terrain. (b) A perspective photo of Devil’s Tower shows just
how steep its sides are. Notice the top of the tower has the same
shape as the center of the topographical map.
Example \(\PageIndex{1}\): Calculating
Equipotential Lines
You have seen the equipotential lines of a point charge in
Figure \(\PageIndex{1}\). How do we calculate them? For example, if
we have a \(+10-nC\) charge at the origin, what are the
equipotential surfaces at which the potential is (a) 100 V, (b) 50
V, (c) 20 V, and (d) 10 V?
Strategy
Set the equation for the potential of a point charge equal to a
constant and solve for the remaining variable(s). Then calculate
values as needed.
Solution
In \(V = k\dfrac{q}{r}\), let V be a constant. The only
remaining variable is r; hence, \(r = k\dfrac{q}{V} =
constant\). Thus, the equipotential surfaces are spheres about the
origin. Their locations are:
This means that equipotential surfaces around a point charge are
spheres of constant radius, as shown earlier, with well-defined
locations.
Example \(\PageIndex{2}\): Potential
Difference between Oppositely Charged Parallel Plates
Two large conducting plates carry equal and opposite charges,
with a surface charge density \(\sigma\) of magnitude \(6.81 \times
10^{-7} C/m\), as shown in Figure \(\PageIndex{8}\). The separation
between the plates is \(l = 6.50 \, mm\).
What is the electric field between the plates?
What is the potential difference between the plates?
What is the distance between equipotential planes which differ
by 100 V?
Figure \(\PageIndex{8}\): The electric field between
oppositely charged parallel plates. A portion is released at the
positive plate.
Strategy
Since the plates are described as “large” and the distance
between them is not, we will approximate each of them as an
infinite plane, and apply the result from Gauss’s law in the
previous chapter.
Use \(\Delta V_{AB} = - \int_A^B \vec{E} \cdot d\vec{l}\).
Since the electric field is constant, find the ratio of 100 V
to the total potential difference; then calculate this fraction of
the distance.
Solution
a. The electric field is directed from the positive
to the negative plate as shown in the figure, and its magnitude is
given by
b. To find the potential difference \(\Delta V\) between the
plates, we use a path from the negative to the positive plate that
is directed against the field. The displacement vector \(d\vec{l}\)
and the electric field \(\vec{E}\) are antiparallel so \(\vec{E}
\cdot d\vec{l} = - E \, dl\). The potential difference between the
positive plate and the negative plate is then
\[ \begin{align*} \Delta V &= - \int E \cdot dl \\[4pt]
&= E \int dl \\[4pt] &= El \\[4pt] &= (7.69 \times 10^4
V/m)(6.50 \times 10^{-3}m) \\[4pt] &= 500 \, V \end{align*}
\]
c. The total potential difference is 500 V, so 1/5 of the
distance between the plates will be the distance between 100-V
potential differences. The distance between the plates is 6.5 mm,
so there will be 1.3 mm between 100-V potential differences.
Significance
You have now seen a numerical calculation of the locations of
equipotentials between two charged parallel plates.
Exercise \(\PageIndex{1}\)
What are the equipotential surfaces for an infinite line
charge?
Answer
infinite cylinders of constant radius, with the line charge as
the axis
Distribution of Charges on Conductors
In Example \(\PageIndex{1}\) with a point charge, we found that
the equipotential surfaces were in the form of spheres, with the
point charge at the center. Given that a conducting sphere in
electrostatic equilibrium is a spherical equipotential surface, we
should expect that we could replace one of the surfaces in Example
\(\PageIndex{2}\) with a conducting sphere and have an identical
solution outside the sphere. Inside will be rather different,
however.
Figure \(\PageIndex{9}\): An isolated conducting
sphere.
To investigate this, consider the isolated conducting sphere of
Figure \(\PageIndex{9}\) that has a radius R and an excess
charge q. To find the electric field both inside and
outside the sphere, note that the sphere is isolated, so its
surface change distribution and the electric field of that
distribution are spherically symmetric. We can therefore represent
the field as \(\vec{E} = E(r)\hat{r}\). To calculate \(E(r)\), we
apply
Gauss’s law over a closed spherical surface S of
radius r that is concentric with the conducting sphere.
Since \(r\) is constant and \(\hat{n} = \hat{r}\) on the
sphere,
\[\begin{align} \oint \vec{E} \cdot \hat{n} \, da &= E(r)
\oint da \\[4pt] &=E(r) 4\pi r^2. \end{align}\]
For \(r < R\), \(S\) is within the conductor, so recall from
our previous study of Gauss’s law that \(q_{enc} = 0\) and Gauss’s
law gives \(E(r) = 0\), as expected inside a conductor at
equilibrium. If \(r > R\), S encloses the conductor so
\(q_{enc} = q\). From Gauss’s law,
\[E(r) 4\pi r^2 = \dfrac{q}{\epsilon_0}.\]
The electric field of the sphere may therefore be written as
As expected, in the region \(r \geq R\), the electric field due
to a charge q placed on an isolated conducting sphere of
radius R is identical to the electric field of a point
charge q located at the center of the sphere.
To find the electric potential inside and outside the sphere,
note that for \(r \geq R\), the potential must be the same as that
of an isolated point charge q located at \(r = 0\),
for two conducting spheres of radii \(R_1\) and \(R_2\), with
surface charge densities \(\sigma_1\) and \(\sigma_2\)
respectively, that are connected by a thin wire, as shown in Figure
\(\PageIndex{10}\). The spheres are sufficiently separated so that
each can be treated as if it were isolated (aside from the wire).
Note that the connection by the wire means that this entire system
must be an equipotential.
Figure \(\PageIndex{10}\): Two conducting spheres are
connected by a thin conducting wire.
We have just seen that the electrical potential at the surface
of an isolated, charged conducting sphere of radius R
is
\[V = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{R}.\]
Now, the spheres are connected by a conductor and are therefore
at the same potential; hence
\[\dfrac{1}{4\pi \epsilon_0} \dfrac{q_1}{R_1} = \dfrac{1}{4\pi
r\epsilon_0} \dfrac{q_2}{R_2},\] and
\[\dfrac{q_1}{R_1} = \dfrac{q_2}{R_2}.\]
The net charge on a conducting sphere and its surface charge
density are related by \(q = \sigma (4\pi R^2)\). Substituting this
equation into the previous one, we find
\[\sigma_1 R_1 = \sigma_2 R_2.\]
Obviously, two spheres connected by a thin wire do not
constitute a typical conductor with a variable radius of curvature.
Nevertheless, this result does at least provide a qualitative idea
of how charge density varies over the surface of a conductor. The
equation indicates that where the radius of curvature is large
(points B and D in \(\PageIndex{11}\)),
\(\sigma\) and E are small.
Similarly, the charges tend to be denser where the curvature of
the surface is greater, as demonstrated by the charge distribution
on oddly shaped metal (Figure \(\PageIndex{11}\)). The surface
charge density is higher at locations with a small radius of
curvature than at locations with a large radius of curvature.
Figure \(\PageIndex{11}\): The surface charge density
and the electric field of a conductor are greater at regions with
smaller radii of curvature.
A practical application of this phenomenon is the
lightning rod,
which is simply a grounded metal rod with a sharp end pointing
upward. As positive charge accumulates in the ground due to a
negatively charged cloud overhead, the electric field around the
sharp point gets very large. When the field reaches a value of
approximately \(3.0 \times 10^6 N/C\) (the dielectric strength of the
air), the free ions in the air are accelerated to such high
energies that their collisions with air molecules actually ionize
the molecules. The resulting free electrons in the air then flow
through the rod to Earth, thereby neutralizing some of the positive
charge. This keeps the electric field between the cloud and the
ground from getting large enough to produce a lightning bolt in the
region around the rod.
An important application of electric fields and equipotential
lines involves the heart. The heart relies on electrical signals to
maintain its rhythm. The movement of electrical signals causes the
chambers of the heart to contract and relax. When a person has a
heart attack, the movement of these electrical signals may be
disturbed. An artificial pacemaker and a defibrillator can be used
to initiate the rhythm of electrical signals. The equipotential
lines around the heart, the thoracic region, and the axis of the
heart are useful ways of monitoring the structure and functions of
the heart. An electrocardiogram (ECG) measures the small electric
signals being generated during the activity of the heart.
PheT
Play around with this simulation
to move point charges around on the playing field and then view the
electric field, voltages, equipotential lines, and more.