Chapter 26 Review Sheet : Capacitance and Dielectrics
Brandon Azad
Summary
This unit covers the concept of a capacitor, a device that stores electric charge. A capacitor is a
combination of two conductors carrying charges of equal magnitude and opposite sign. Energy is stored
in the electric field between the opposing charges of a capacitor.
Terms and Formulae
A capacitor is a combination of two conductors that have equal magnitude charges of opposite sign.
Between the two conductors there is an electric field and, therefore, a potential difference. It is customary
to refer to the charge of a capacitor, which corresponds not to the net charge but the magnitude of charge
on a single conductor. Note that the net charge on a capacitor is zero, regardless of the charge on either of
the conductors.
A quantity called capacitance can be used to describe a capacitor. The
charge Q on either conductor of the capacitor is proportional to the potential
difference between them for that capacitor. This relationship can be
expressed as Q = C ∆V, where C is a proportionality constant called the
capacitance. The capacitance for a capacitor is thus defined as:
C≡
Q
ΔV
The units of capacitance are coulombs per volt, defined as the farad.
Typically, most capacitance values are measured in microfarads or
picofarads.
One common type of capacitor is called a parallel plate capacitor, which consists of two parallel plates
of equal area A separated by some distance d. The capacitance is proportional to A and inversely
proportional to d, and can be described by the equation:
C =ε0
A
d
In this equation, ε0 is the permittivity of free space.
The circuit symbol for a non-polarized capacitor is a set of two parallel bars
perpendicular to the circuit wires.
The equivalent capacitance of a set of capacitors arranged in parallel is the sum of each individual
capacitor's capacitance. Thus, for a set of n capacitors C0, C1, …, Cn
arranged in parallel, the equivalent capacitance is given by:
n
C eq=∑ C k =C 0+C 1+...+C n
k=0
The equivalent capacitance then is larger than the capacitance of any
one capacitor in the combination. This relationship can be determined by recognizing that the potential
difference across each capacitor in the set is the same and that the total charge on the capacitors is the sum
of the charges of each capacitor.
The equivalent capacitance of a set of capacitors arranged in series is the reciprocal of the sum of the
reciprocals of the individual capacitances. This is best explained for a set of n capacitors C0, C1, …, Cn
arranged in series by the relationship:
n
1
1
1
1
1
=∑
= + +...+
C eq k =0 C k C 0 C 1
Cn
The equivalent capacitance then is smaller than the capacitance of any one capacitor in the combination.
This relationship can be determined by recognizing that the sum of the potential differences across each
capacitor in the set is equal to the potential difference across the set and that the charge on each capacitor
is the same.
Because an electric field exists between the plates of a charged capacitor, a capacitor can be used to store
energy. The energy stored in a capacitor is proportional to the charge on each plate and the voltage
difference between the plates:
1
U c= Q Δ V
2
Expressed in terms of capacitance and voltage, the energy stored in a capacitor is:
1
U c= C Δ V 2
2
The energy density in an electric field at a point is proportional to the square of the magnitude of the
electric field:
1
u E= ε0 E 2
2
A dielectric is a nonconducting material. The presence of a dielectric between the plates of a capacitor
increases the capacitance; when the gap between the plates is completely filled with a dielectric, the
capacitance increases by the dielectric constant κ for that material:
C =κC 0
In this equation, C represents the capacitance with the dielectric in place and C0 represents the capacitance
without the dielectric.
The dielectric strength of a dielectric is the maximum electric field that can be set up within the material
before it begins to break down and conduct. Because dielectrics are only insulators up to some electric
field strength, physical capacitors have a maximum voltage rating called the working voltage, breakdown
voltage, or rated voltage.
Problems
1. [Easy] What is the equivalent capacitance of the following network of capacitors?
2. [Medium] Show that the capacitance of a conducting cylindrical rod of radius a and charge Q
coaxial to a conducting cylindrical shell of negligible thickness and charge -Q and radius b such
that b > a, both with length l, is given by:
C=
2 π ε0 l
ln (b/ a)
3. [Hard] The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor's
inside radius is 3.00 mm. The space between the conductors is filled with polyethylene, which has
a dielectric constant of 2.30 and a dielectric strength of 18.0x106 V/m. What is the maximum
potential difference that this cable can withstand?
Solutions
1. To find the equivalent capacitance we apply the rules of circuit simplification. The capacitance of
the topmost branch is simply 3 mF, because there is a single capacitor. That of the second branch
must be calculated using the series equation: the total capacitance is the sum of the individual
capacitances, or 1 mF + 2 mF + 3 mF = 6 mF. Finally, the equivalent capacitance of the lower
branch is 2 mF + 2 mF or 4 mF. To calculate the equivalent capacitance of the whole array, use
the parallel capacitors equation:
1
1
1
1
1
4
=
+
+
=
or C eq = mF =1.33 mF
C eq 3 mF 6 mF 4mF 4/ 3 mF
3
2. To calculate capacitance by C = Q / ∆V, we need to know the potential difference between the
two conductors of the capacitor. This difference can be derived by examining the electric field in
the space between the conductors. The electric field can be derived using Gauss's Law, choosing
as the Gaussian surface a cylindrical shell coaxial to the center rod with radius r > a.
∯S E⋅dA= Qε0
Because the electric field is perpendicular to the surface at all points, the dot product reduces to
the product of the magnitudes. Note that the symmetry assumptions do not hold for the ends of
the rod and Gaussian surface, but we will assume they do.
∫ E dA= EA= Qε0
Solving for E and substituting the surface area of the body of the cylinder for A yields the
magnitude of the electric field at any distance r from the axis of the rod.
E=
Q
Q
=
ε0 A ε 0 2π r l
Using this result, we can integrate along the path of the electric field to determine the potential
difference between the two conductors.
Δ V =−∫ E⋅dr
The direction of the electric field is radially outward from the axis of the rod, meaning the dot
product reduces to the multiplication of the magnitudes of the vectors.
b
Δ V =−∫ E dr=−∫
a
Q dr
2 πε0 l r
b
ΔV=
()
−Q
dr
−Q
b
=
ln
∫
2 π ε0 l a r 2 πε 0 l
a
Substituting the magnitude of the voltage difference into the definition of capacitance yields the
capacitance.
C=
Q 2π ε0 l
2 π ε0 l
Q
=
=
ΔV −Q ln (b/a) ln (b/a)
3. We need to determine the potential difference between the two wires that will cause the electric
field at some point to match the dielectric strength of polyethylene. From the intermediate results
of 2 above, we can examine the relationship between the voltage difference and the electric field
for a conducting cylindrical rod coaxial to a conducting cylindrical shell (the model for a coaxial
cable).
E=
Q
Q 1
=
ε0 2π r l 2 π ε0 l r
From this equation it is evident that the electric field is strongest when r is as small as possible, or
at the surface of the inner cable. Substitute a for r.
E max =
Q 1
or
2 π ε0 l a
E max a=
Q
2 πε0 l
Substitute this result into an intermediate for the potential difference between the inner and outer
cylinders to express the voltage.
()
Q
b
ln
2 π ε0 l
a
Δ V max = E max a ln (b/a)
ΔV=
Substitute the known values of Emax = 18.0x106 V/m, a = 0.800 mm = 8.00x10-4 m, and b = 3.00
mm = 3.00x10-3 m.
Δ V max =(18.0×106 V /m)⋅(8.00×10−4 m)⋅ln
(
)
(3.00×10−3 m)
=19.0 kV
(8.00×10−4 m)