1
Handout 4: Capacitance and dielectrics
Capacitance
+𝑄
−𝑄
Capacitance 𝐶 refers the ability of an object to
store charge. In Figure 1, voltage 𝑉 is supplied and the
charge 𝑄 is stored. The capacitance is defined as
𝑉
𝑄
𝐶 = .
𝑉
The unit of 𝐶 is coulomb per volt or farad (F). Figure 2
shows a linear relationship between the charge stored
and voltage. The slope is equal to the capacitance. A
device that stores charge is called capacitor.
Figure 1: Voltage is supplied to
store charge
In Figure 1, to store more charge, a small charge
𝑑𝑄 must be moved from – 𝑄 to +𝑄 while voltage is V.
Work done is equal to the energy stored in the capacitor:
𝑄
𝑊 = 𝑈 =
𝑄
𝑉 𝑑𝑄 =
0
0
𝑄
𝑑𝑄
𝐶

𝑄2
𝑈 =
.
2𝐶
One can use equation 𝑄 = 𝐶𝑉 to obtain the other
expressions for the energy stored:
𝑈 =
1 2
𝐶𝑉
2
and
𝑈 =
1
𝑄𝑉.
2
The energy stored is the area under graph in Figure 2.
Example 1 Calculate the capacitance of the spherical
conductor of radius 𝑅 = 0.25 m.
Example 2 A defibrillator, a device for treatment
cardiac arrest, consists of a capacitor 𝐶 connecting
two electrodes. If C = 300 𝜇F and energy of 300 J is
be passed from the capacitor, how much voltage
required?
of
to
to
is
Figure 2: Graph of charge against
voltage. The slope is the capacitance
2
Parallel-plate capacitor
Figure 3 shows a parallel-plate capacitor. If the
area of the plate is 𝐴, the electric field between the
plates is 𝐸 = 𝜎 𝜀0 = 𝑄 𝜀0 𝐴. Substituting E into the the
expression for voltage 𝑉 = 𝐸𝑑 leads to 𝑉 = 𝑄 𝜀0 𝐴 𝑑.
Therefore, the capacitance is given by
𝐶 =
𝑄
𝜀0 𝐴
=
.
𝑉
𝑑
Figure 3: Parallel-plate capacitor
The capacitance of the parallel-plate capacitor depends
on area 𝐴 and separation 𝑑 of the plates.
Another is quantity of interest is the energy
stored in the capacity, which is often expressed as
energy density (energy per unit volume) 𝑢:
𝑢 =
𝑈
1
1 𝜀0 𝐴
=
𝐶𝑉 2 =
𝐴𝑑
2𝐴𝑑
2𝐴𝑑 𝑑
𝐸𝑑 2 .
Hence, the energy density
𝑢 =
1
𝜀0 𝐸 2 .
2
The is the general equation for energy density stored in
electric field E, valid for any shape of capacitor.
Spherical capacitor
Figure 4 shows a spherical capacitor consisting of
two concentric spheres containing equal magnitude of
charge Q but opposite signs. The inner and the outer
radii are 𝑎 and 𝑏 respectively. Electric field in the space
in the capacitor is 𝐸 𝑟 = 𝑄 4𝜋𝜀0 𝑟 2 . From the equation
𝐸 = − 𝑑𝑉 𝑑𝑟,
𝑏
−
𝑏
𝑑𝑉 =
𝑎
𝑎
𝑄
𝐸𝑑𝑟 =
4𝜋𝜀0
𝑏
𝑎
1
𝑑𝑟.
𝑟2
The term on the left is the voltage 𝑉 = 𝑉 𝑎 − 𝑉 𝑏
which is given by
𝑉 =
𝑄 1 1
𝑄
𝑏−𝑎
−
=
.
4𝜋𝜀0 𝑎 𝑏
4𝜋𝜀0
𝑎𝑏
The capacitance,
𝐶 =
𝑄
𝑎𝑏
= 4𝜋𝜀0
.
𝑉
𝑏−𝑎
Figure 4: Spherical capacitor
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Cylindrical capacitor
Figure 5 shows a cylindrical capacitor consisting of
two coaxial cylinders of length ℓ containing equal
magnitude of charge Q but opposite signs. The inner and
the outer radii are 𝑎 and 𝑏 respectively. By using Gauss’
law, the electric field between the cylinders is
𝐸 𝑟 = 𝜆 2𝜋𝜀0 𝑟, where 𝜆 = 𝑄 ℓ is the charge per unit
length. By integrating 𝐸 = − 𝑑𝑉 𝑑𝑟,
𝑏
−
𝑏
𝑑𝑉 =
𝑎
𝑎
𝜆
𝐸𝑑𝑟 =
2𝜋𝜀0
𝑏
𝑎
1
𝑑𝑟.
𝑟
The term on the left is the voltage 𝑉 = 𝑉 𝑎 − 𝑉 𝑏 :
𝑉 =
Figure 5: Cylindrical capacitor
𝜆
𝑏
𝑄
𝑏
ln =
ln .
2𝜋𝜀0 𝑎
2𝜋𝜀0 ℓ 𝑎
The capacitance is thus
𝐶 =
𝑄
2𝜋𝜀0 ℓ
=
.
𝑏
𝑉
ln
𝑎
Connecting capacitors
Consider Figure 6 where capacitors are connected
in series. Equal amount of charge 𝑄 is induced on all
capacitors. The total charge stored in also 𝑄. The total
amount of voltage 𝑉 is divided into two parts 𝑉1 and 𝑉2 :
𝑉 = 𝑉1 + 𝑉2

𝑄
𝑄 𝑄
=
+ .
𝐶
𝐶1 𝐶2
The equation for the total capacitance 𝐶 is
1
1
1
=
+ .
𝐶
𝐶1 𝐶2
Figure 6: Capacitors in series
(series)
In Figure 7, capacitors are connected in parallel. In
this case the voltage 𝑉 is equal for the capacitors but the
total charge 𝑄is split into 𝑄1 and 𝑄2 :
𝑄 = 𝑄1 + 𝑄2

𝐶𝑉 = 𝐶1 𝑉 + 𝐶1 𝑉.
The equation for the total capacitance 𝐶 is
𝐶 = 𝐶1 + 𝐶2 .
(parallel)
Figure 7: Capacitors in parallel
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Example 3 Three capacitors in the Figure are connected
in parallel between to the terminals A and B which are
connected to 10-V voltage.
a) Find the total capacitance.
b) Determine the total charge stored in the circuit.
c) Find the charge stored in individual capacitors.
Example 4 Find the total capacitance of the following
combination of capacitors.
Capacitor with dielectrics
Dielectric, an insulating material such as oil or
plastic, is inserted between the plates of a capacitor in
order to increase the capacitance of the capacitor. With a
dielectric completely fills the space between the plates
(Figure 8), the capacitance is increased by a factor 𝜅
called the “dielectric constant”:
Figure 8: Dielectric is inserted
between the plates of a capacitor
𝐶 = 𝜅𝐶0 ,
where 𝐶0 is the capacitance before the dielectric is
inserted. The values of 𝜅 of different materials are
shown in Figure 9.
Dielectric is insulator. When the applied electric
field is large enough, the dielectric material breaks down
and forms a conducting path between the plates. The
maximum value of electric field that the dielectric
material can tolerate before breaking down is called
“dielectric strength” (Figure 9).
Figure 9: Dielectric constants
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Consider an isolated capacitor (constant charge 𝑄)
in Figure 10(a). Since 𝐶 is increased by a factor of 𝜅.
From 𝐶 = 𝑄/𝑉, the voltage is reduced by the same factor
of 𝜅, i.e., 𝑉 = 𝑉0 /𝜅. On the contrary, in Figure 10 (b), the
capacitor is connected to a voltage source. Hence, the
voltage is constant. From 𝐶 = 𝑄/𝑉, if 𝐶 is increased by a
factor of 𝜅, the charge stored 𝑄 is also increased by the
same factor, i.e., 𝑄 = 𝜅𝑄0 .
Figure 11 shows an isolated parallel-plate
capacitor with a dielectric inserted between the plates.
Negative charge is induced near the positive plate while
positive charge is accumulated near the negative plate.
As a result, there is an induced electric field 𝐸𝑖 in the
dielectric opposing the field 𝐸0 from the metal plates.
Therefore, the net electric field is given by
(a)
(b)
Figure 10: (a) Isolated charged capacitor
and (b) capacitor in circuit
𝐸 = 𝐸0 − 𝐸𝑖 .
The electric field inside the plates is reduced due to the
presence of dieclectric. For isolated capacitor, the
voltage is reduced by a factor 𝜅. From 𝑉 = 𝐸𝑑, the
electric field is also reduced by the same factor:
𝐸 =
Therefore,
𝐸0
= 𝐸0 − 𝐸𝑖
𝜅
𝐸0
.
𝜅
⇒
𝐸𝑖 =
1−
1
𝐸.
𝜅 0
Figure 11: Dielectric is inserted
between the plates of a capacitor

The above equation expresses the induced field 𝐸𝑖 in
terms of the original field 𝐸0 of an isolated capacitor.
From Gauss’ law, the field between the plates is
given by 𝐸 = 𝜎 𝜀0 . From , we obtain the relationship
between the induced charge 𝜎𝑖 and the free charge 𝜎0 :
𝜎𝑖 =
1−
1
𝜎.
𝜅 0
Example 5 An isolated parallel-plate capacitor, with
capacitance 𝐶0 = 5.0 𝜇F is charged with a voltage
𝑉0 = 10 V. A material with dielectric constant 𝜅 = 2.2 is
completely filled between the plates. How much energy
is stored in the dielectric-filled capacitor?
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Example 6 Two parallel palates of area 𝐴 = 100 cm2 are
given charges of equal magnitudes 8.4 × 10−7 C but
opposite signs. The electric field within the dielectric
filling the space between the plates is 𝐸 = 1.4 MVm-1.
a) Calculate the dielectric constant of the material.
b) Determine the magnitude of the charge induced
on each dielectric surface.
*Example 7 A parallel-plate capacitor with plate area 𝐴
and separation 𝑑 is filled with two kinds of dielectric
materials with constants 𝜅1 and 𝜅2 as shown in the
diagram (a) and (b). Determine the expressions for the
capacitance in (a) and (b)
*Example 8 The parallel plates of a capacitor are given
equal free charge of density 𝜎 but opposite signs. The
plates are separated by distance 𝑑. A slab of dielectric
material with constant 𝜅 and thickness 𝑎 is inserted
between the plates.
Show that the voltage between the plates is given by
𝑉 =
𝜎
1
𝑑− 1− 𝑎 .
𝜀0
𝜅