C
T
=
C
1
C
2
C
1
+C
2
C
T
= ⌈
3×6
3+6
⌉μF
C
T
=
18
9
μF
C
T
=2 μF
Example 02
The figure below shows a network of
capacitors. Find the charge in the 1.5μF
capacitor.
Solution
Step 1: Consider the circuit of 2μF and 3μF
capacitors. The two are in parallel, therefore.
C
p
= C
1
+C
2
C
p
= 2μF+3μF
C
p
= 5μF
Step 2: Join the 5μF and the 1.5μF together,
and the new circuit will be in series. Then:
C
T
=
C
1
C
2
C
1
+C
2
C
T
= ⌈
5×1.5
5+1.5
⌉μF
C
T
=
7.5
6.5
μF
C
T
=1.15 μF
Step 3: To find the charge, Q in the 1.5μF
capacitor
Since the 5μF and the 1.5μF are in series
they have the same charge which the total
charge flowing in the circuit.
Total capacitance, C= 1.15μF=1.15× 10
-6
F
Charge, Q is given by
Q= C×V
= 1.15× 10
-6
F×20V
= 2.3× 10
-5
C
The total charge flowing through 1.5μF
capacitors is 2.3× 10
-5
C
Example 03
Three capacitors of capacitance values
3μF, 4μF and 5μF are connected as shown
in the figure below. Study it carefully and
answer the next questions.
Calculate the charge stored in the 5μF
capacitor
Solution
Step 1: Consider the circuit of 4μF and 5μF
capacitors. The two are in parallel, therefore.
C
p
= C
1
+C
2
C
p
= 4μF+5μF
C
p
= 9μF
Step 2: Join the 9μF and the 3μF together,
and the new circuit will be in series. Then:
C
T
=
C
1
C
2
C
1
+C
2
C
T
= ⌈
9×3
9+3
⌉μF
C
T
=
27
12
μF
C
T
=2.25 μF
Step 3: To find the total charge
Q= C×V
= 2.25× 10
-6
F×10V
= 2.25× 10
-5
C
Step 4: To find the potential in the parallel
circuit, V
2
Solution
Total capacitance, C= 2.25μF=2.25× 10
-6
F
Charge, Q is given by
V
2
=
charge
capacitance
V
2
=
2.25× 10
-5
C
9× 10
-6
F
V
2
= 2.5V
Step 5: Charge on the 5μF capacitor
Since the 4μF and 5μF are in parallel, the
p.d is the same.
Q= C×V
= 5× 10
-6
F×2.5V
= 1.25× 10
-5
C
FACTORS AFFECTING THE
CAPACITANCE
Capacitance of a parallel plate capacitor is
affected by three major factors, namely:
(i) The area of the plate
(ii) The insulating property (dielectric) of
the medium
(iii)
The distance between the plates
Area of the plate
An increase in the area of the plates causes a
decrease in potential difference between the
plates, hence the capacitance increases.
capacitance
(
C
)
∝area (A)
The insulating property of the medium
Capacitance depends on the dielectric
(insulating) material between the plates.
Example if you use glass, paper or polythene
rod instead of air, the capacitance increases
capacitance
(
C
)
∝dielectric constant(ε)
The distance between the plates
The capacitance increases if the distance
between the plates decreases.
capacitance∝
1
d
Example 01
Give three ways on how you can do to
increase the capacitance of the parallel plate
capacitor.
Answer
(i) Reducing the distance of separation of
the plates
(ii) Increasing the area of the plates
(iii) Using a good insulating (dielectric)
material.
Example 02
Briefly explain the factors affecting the
capacitance of the capacitor
Answer
(i) Area of the plates: Increasing the area
of the plates increases the capacitance
of the capacitor.
capacitance
(
C
)
∝area (A)
(ii) Distance of separation between the
plates: The increase in distance of
separation between the plates reduces
the capacitance of the capacitor
capacitance∝
1
d
(iii) Insulating property (dielectric) of the
medium: The capacitance depends on
the insulating property of the medium,
example plastic increases the
capacitance compared to air
capacitance
(
C
)
∝dielectric constant(ε)
Example 03
What happens to the capacitance of the
capacitor when
(a) The area of the plate increases
Answer
The capacitance of the capacitor will
increase
(b) The distance between the plates
decreases
Answer
The capacitance of the capacitor will
increase
(c) Potential difference between the
plates decreases
Answer
The capacitance of the capacitor will
increase
3.40 CHARGE DISTRIBUTION
Charge distribution on a conductor is mostly
concentrated at sharply curved surfaces.
To investigate the charge distribution on a
conductor, there are two devices we can use.
(i) A proof plane
(ii) A leaf electroscope
A proof plane is a metal disk with an
insulated handle through its disc used to
transfer small amount of charge from one
body to another.
Action of a proof plane
A proof plane is touched to the surface of a
conductor at various points in turn. The
charge on the proof plane is then transferred
to an electroscope which indicates the
charge distribution on the conductor.
Note that: The charge distribution on the
conductor depends on the shape of the
conductor as explained below.
Charge distribution on a spherical
conductor
Spherical conductor has equal distribution of
charges. So the leaf diverge is the same
throughout.
Charge distribution on a pear – shaped
conductor
More charges concentrate at the sharp point
as shown below.
Charge distribution on a rectangular
conductor
Charges concentrate at the sharp points (at
the corners) of the conductor as shown
below.
Charge distribution on a hollow
conductor
In a hollow conductor, charges always
reside outside the conductor. The inside of
the can has no charge.
FARADAY’S ICE-PAIL EXPERIMENT
This is an experiment to shoe charge
distribution in a hollow conductor
Uncharged metal ice pail is placed on the
uncharged gold leaf electroscope. A
positively charged metal sphere is
suspended on a thread and lowered it into
the pail without touching the pail.
Observation: The gold leaf diverges
indicating that charge has been induced on
the outside of the pail.
