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CHAPTER 5
ELECTRIC POTENTIAL AND CAPACITANCE
Example of electric field in real application:
1. A flow of electric charge when a spark jumps across a small distance in air.
(Lightning in the sky-there exist a high potential during a thunderstorm)
2. Spark plug
3. Parallel plate capacitor
4. Electro Cardiogram (ECG)
Unit for Electric Potential is Volt (V) in honor of Sir Alessandro Volta, and Italian
Scientist (1745-1827)
Electric Potential Energy Difference
Consider that you move the positive charge form the negative plate to the
positive plate. Work has to be done to go against the electric field. If the force is
released, the proton will move back to plate A. The amount of work done is
define as
Ue =UB- UA = qoEd
Consider you move a box of mass m at a height h, so the potential energy
Ug =UB- UA = Fh = mgh
Electric Field Difference (Volt) For Parallel plate
In dealing with electric forces, we eliminated the dependence on the test charge
by defining the electric field as the electric force per unit charge. By knowing the
electric field, we could determine the force on any point charge place in that field
(From Fe= qoE).
Similarly, we define the Electric Potential Difference (V) between any two points
in space as the change in potential energy per unit positive test charge. In term
of the work done on the electric charge by an external force
V=
U o W

qo
qo
SI unit for electric potential difference: joule/ coulomb (J/C) or volt (V)
V=
U o qo Ed

 Ed
qo
qo
Consider a test charge is move from position A to B
For Point Charge:
Work is define is force times distance
W = F. r
W=
W=
1
.r
4 o r 2
1
4 o r
Qq
For a general case W = kQqo (
1 1
 )
rB rA
If you consider rB is very far away, (rB = ) than
1
PE electric = W = kQqo ( )
r
For comparison
PE gravitation = -
VAB = VB – VA =
W AB
q
= k
r
q
GMm
r
or WAB = qVAB
Always remember the law regarding to conservation of energy
KEi + PE i = KEf + PE f
Initial energy = final energy
Example:
1. A device consists of two parallel plates separated by a distance of 0.300
mm. If a 20.0-V potential difference is maintained between those plates,
calculate the electric field strength in the region between the plates.
ANS:

 
V = –Ed = –E(3.00  10–4 m) = –20.0 V
so E = 6.67  104 V/m
2. A proton is between two plates, separated by a distance of 2.0 cm, across
which there is a potential difference of 5000 V. (a) Find the electrical force
exerted on the proton, and (b) its acceleration.
Ans:
(a) The strength of the electric field is
E=
V
5.0  103 V
=
= 2.5  105 N/C
d
2.0  10–2 m
and F = qE = (1.60  10–19 C)(2.5  105 N/C) = 4.0  10–14 N
(b) a =
F
4.0  10–14 N
=
= 2.4  1013 m/s2
m 1.67  10–27 kg
Potential Difference Due to Point Charges
The expression for the potential difference (voltage) between two points at a
distance rA and rB from a point charge (q) is
V=
kq kq
electric potential difference (point charge only)

rB rA
If B is closer to the point charge than A, the potential difference is positive: B is at
a higher potential than A.
Electric potential increases (+V) as we consider location nearer to positive
charges or farther from negative charges.
Electric potential decreases (-V) as we consider location farther from positive
charges or near to negative charges.
The potential at a very large distance from a point charge is usually defined to be
zero; the reference point is convenient but not required. For this choice of
reference, the electric potential at a distance r from the charge is
V=
kq
r
Example:
1. At what distance from a point charge of 6.0 µC would the potential equal
2.7 x 104 V?
Ans:
V=k
q
kq (8.99  109 N m2/C2)(6.0  10–6 C)
, or r =
=
= 2.0 m
r
V
2.7  104 N m/C
2. In the Bohr model of the hydrogen atom an electron circles a proton in an
orbit of radius
0.510 x 10–10 m. Find the potential at the position of the electron.
Ans:
V=k
q
(8.99  109 N m2/C2)(1.60  10–19 C)
=
= 28.2 V
r
0.510  10–10 m
The Van de Graaff Electrostatic Generator
An electrostatic generator is a device for producing very large electric potential
difference, of the order of millions of volts.
It was develop in 1931 by the American physicist Robert J. Van de Graaff (19011967).
The main application for these electrostatic generators is to accelerate charged
particles to a very high kinetic energy.
Example:
What is the speed of a singly charged boron ion (in an ion implanter) that has
been accelerated from rest by a potential difference of 86 kV?
Ans:
Our assumption Kinetic Energy = Potential Energy
½ mB v2 = q V
v=
v=
2qv
mB
2(1.6 x10  19 C )(86 x10 3 V )
1.83x10 26 kg
v = 1.23 x 106 m/s
The Electron Volt (A unit of energy)
We define Voltage (V) =
W
so W= qV= e(1V) = (1.602 x 10-19C) (1V)
q
= 1.602 x 10 –19J
1 eV = 1.602 x 10-19 J
Equipotential Surface.
Any surface for which all points are at the same potential is called an
Equipotential surface.
When all charges have come to rest, the surface of the conductor is always an
Equipotential surface.
Capacitors
Capacitor is a device for storing charges.
C=
q
V
C is measured is farad (F) in honor of English Physicist and chemist Michael
Faraday (1791 –1867)
1 farad = 1 Coulomb / Volt
Parallel Plate Capacitor
A capacitor can be made by putting two metal plate separated by an insulator or
dielectric
C=
A=
Cd
o
A o
d
where  o = 8.85 x 10–12 F.m-1
Example:
1. To what difference in potential would a 2.0-µF capacitor have to be
connected in order to store 98 µC of charge?
Ans:
V =
Q 98 C
=
= 49 V
C 2.0 F
2. If the area of the plates of a parallel plate capacitor is doubled while the
spacing between the plates is halved, how is the capacitance affected?
C=
 oA
d
, so if A2 = 2A1 and d2 =
1
d , then
2 1
C2 =
 oA 2
d2
=
o2A1
1
d
2 1
=4
 oA 1
d1
or C2 = 4C1 (The capacitance is quadrupled.)
Dielectric
When a non conducting (electrically insulating) material known as dielectric is
inserted between the plates, the capacitance increases even though the
separation remains constant
Dielectric constant  (Kappa)
=
C (dielectric layer) q / V Vo

=
Co (Vacuum layer) q / Vo V
C=
 o A
d
Energy stored in a capacitor.
Capacitor is device that can store charge. The charge is retain in its electric field.
So, a capacitor has a capability to store energy.
Q (Charge)
Voltage (V)
Energy store = ½ q V
q=CV
Energy stored = ½ CV2
Example:
1. (a) A 3.00-µF capacitor is connected to a 12.0-V battery. How much
energy is stored in the capacitor? (b) If the capacitor had been connected
to a 6.00-V battery how much energy would have been stored?
Ans:
(a) W =
1
1
CV2 = (3.00 F)(12.0 V) 2 = 2.16  10–4 J
2
2
(b) W =
1
1
CV2 = (3.00 F)(6.00 V) 2 = 5.40  10–5 J
2
2
CAPACITOR IN SERIES
For capacitor in series:
All the charge is the same anywhere in the capacitor, where QT=Q1=Q2=Q3
C1
C2
C3
1
1
1
1



CT C1 C 2 C3
CAPACITOR IN PARALLEL
For capacitor in parallel, the charge store in the total capacitor is QT=Q1+Q2+Q3
The total Capacitance CT = C1+C2+C3
C1
C2
C3