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Equipotential Surfaces and Capacitors
Definition An equipotential surface is a locus of points that are all at the same electric potential.
Thus an equipotential surface is a surface on which the electric potential is the same everywhere. The
equipotential surfaces surrounding an isolated point charge are concentric spheres. See the figure below.
Taking the electric potential to be zero at infinity, the electric potential at
point D in the figure is
kq
r
. The electric potential at point B is lower than
that at point D because point B is farther away from the charge q.
If a small positive test charge q 0
moves along the outer equipotential
surface along the path from A to B the work W AB
done by the electric field
is given by Equation 19.4:
V
B
– V
A
= –W
AB
/q
0
But V B
= V
A
so that W AB
= 0 and the electric field does no work on the test
charge. This is possible only if the electric force acts in a direction that is
perpendicular to the displacement of the test charge. (Remember the
formula for work from Physics 111: W = Fs cos(), where is the angle
between the force F and the displacement s. If = 90 then W = 0 J.)
We conclude that the electric field lines intersect an equipotential surface at right angles. Note that the electric
field lines are directed from high electric potential to low electric potential.
Question: In the figure at the right, which of the three equipotential surfaces shown
is at the highest electric potential? Which is at the lowest?
There is a quantitative relationship between the electric field and equipotential
surfaces. Consider a region of space where there is a uniform electric field directed
from left to right. (Such a region exists, for example, near the center of a parallel-
plate capacitor. See the figure below.) In this region the equipotential surfaces are
parallel vertical planes.
If a small positive test charge q
0
moves from
point A to point B the work done on the charge by
the electric field is given by
W
AB
= F s
Where F is the magnitude of the electric force on q 0
F = q 0
E
E
E
Thus
W
AB
= q 0
E s
But from Equation 19.4 V B
– V
A
= – W
AB
/q 0
so that
V
B
– V
A
= E s
or
E = V / s
where V = V B
– V
A
. The quantity V / s is referred to as the potential gradient and has units of volts per meter.
Remarks. The potential gradient is a vector field that points in the direction of the most rapid rate of increase of
the potential. Because of the negative sign in the above formula we see that the electric field points in the
direction of the most rapid rate of decrease of the potential. (Proofs of these statements are beyond the scope of
this course. However, interested students may refer to a suitable calculus text such as Anton’s Calculus .)
Remark. Because of the above formula we see that alternate units for the electric field are volts per meter : V/m.
Exercise: Show that 1 V/m = 1 N/C.
Example
The plates of a parallel plate air capacitor are connected to the terminals of a 5.0 volt battery. The plates are
separated by a distance of 3.0 mm. Neglect fringing effects.
a. Find the magnitude of the electric field between the plates.
b. Find the surface charge density that must exist on each plate.
a.
3
3 V
m 3
; 5.0 V; 3.0 10 m
5.0 V
3.0 10 m
V
E V s
s
E E
b.
2
2
2 2
0
12 3 C V
0 0 m
N m
12 3 8 C C N C
C N m m
For a parallel plate air capacitor , where is the surface charge density on either plate. Thus
E
E E
Since E 0
E the dielectric constant is a number greater than unity. See Table 19-1 in your text for dielectric
constants of some common substances.
TABLE 19.1 Dielectric Constants of Some Common Substances
a
Substance Dielectric Constant, k
Vacuum 1
Air 1.
Teflon 2.
Benzene 2.
Paper (royal gray) 3.
Ruby mica 5.
Neoprene rubber 6.
Methyl alcohol 33.
Water 80.
The Capacitance of a Parallel Plate Capacitor
Recall from the previous example that the magnitude of the electric field between the plates of a parallel plate
capacitor can be calculated from
or where and is the plate separation.
V V
E E V V d s
s d
Now E = E 0
/ and E 0
0,
where is the surface charge density on either plate: = q / A where A is the
surface area of each plate. Thus
0
0 0 0
0
0
or
E V V V V q V
E
d d d d A d
A
q V
d
A comparison of the last equation with q = CV shows that for a parallel plate capacitor
0
A
C
d
Note that from last formula above we see that 0
can be expressed in the units F/m. (Why?)
Example
Suppose an air capacitor consisted of two identical square copper plates placed side by side with a plate
separation of 1.0 cm. Find the cross-sectional area of each plate if the capacitance of the capacitor is to be 1.0 F.
0 0
0
2
9 2
12 F
m
; here 1 so that and.
1.0 F 1.0 10 m
; 1.1 10 m
A A Cd
C C A
d d
A A
Each of these plates would have a side over 33 km long! This example illustrates that, typically , common
capacitors that you hold in your hand have capacitances that are only small fractions of a farad.
Effect of a Dielectric on Capacitance
From the previous example we saw that if just air is between the plates of a parallel plate capacitor the
capacitance is
0 0
A
C
d
. When a dielectric with dielectric constant is placed between the plates the
capacitance becomes
0
A
C
d
. We can conclude that, for the parallel plate capacitor,
0
C C
It can, in fact, be shown that this formula is valid for all capacitors, regardless of their shape. So the insertion of
a dielectric between the plates of a capacitor has the effect of increasing the capacitance of the capacitor.
Remark. The use of a dielectric to increase capacitance has a drawback. When a dielectric is subjected to an
electric field that is strong enough the dielectric breaks down and becomes a conductor. The minimum electric
field strength to which a dielectric can be subjected without breakdown is called the dielectric strength of the
dielectric. The dielectric strength of air, for example, is 3 10
6
V/m; for paper it is 16 10
6
V/m. Placing a
dielectric between the plates of a capacitor, while increasing the capacitance, also limits the potential difference
that can be placed across the plates of the capacitor. If too high a potential difference is placed across the
capacitor’s plates the dielectric strength of the dielectric between the plates can be exceeded causing the
capacitor to break down and become useless.
Synthetic dielectrics have been developed that have enormous dielectric constants. A demonstration will be
given in class of capacitor with a cross-sectional area about the size of a quarter that has a capacitance of 1.0 F.
Example
a. Estimate the capacitance of a capacitor made from two quarters with a piece of notebook paper placed
between them.
b. What would the dielectric constant of the material placed between the quarters have to be to produce a
capacitance of 1.00 F? (Assume the same separation of the quarters.)
