Lecture 3.8
ELECTRICITY
Potential Difference
Capacitance
Electric current
Electrical Potential Difference
more commonly known as voltage (V)
e.g. 9V battery
Analogy:
Open tube filled with water
Equal height:
Equal potential energy
Higher
potential energy
No water flow
Lower potential
energy
Battery
chemical reactions produce electrical
potential difference between terminals
electrons to flow in external circuit
Electrical Potential Difference
Definition of voltage
Q
+++ +
+ +++
+ ++
+
Test charge
+q
+
F
A
B
∆r
Test charge experiences a repulsive Coulomb
force, therefore it has
electrical potential energy due to its position
Test charge free to move from A to B,
• Electric field does work via Coulomb force
• potential energy decreases
Work done on test charge is
WAB= F ∆r
where F is the average force on charge +q
For the expert: For
point charge, force
decreases with
=
WAB
distance. For
small distances:
B
∫ F (r )dr ≈ F ∆r
A
Electrical Potential Difference
Hence potential energy (PE) ∆U of test charge
decreases by WAB in going from A to B.
Change in PE,
∆U = -WAB
1 Qq
F=
4πε 0 r 2
Coulomb force F
therefore
∝q
WAB ∝ q
∆U ∝ q
Voltage between A and B is defined as
the change in electric potential energy
as charge q moves from A to B
VAB
∆U
=
q
U
V=
q
V is the potential energy per unit charge
Electrical Potential Difference
Q
+++ +
+ +++
+ ++
+
A
Note: Electric Potential (due to Q) exists at
point A even if there is no test charge q there.
Voltage is defined as:
•potential energy per unit charge
•or potential difference
SI unit for voltage is the volt (V)
U
V=
q
1V = 1J/C
Volt: named after Alessandro Volta (1745-1827),
Italian scientist who invented the battery.
Voltage of a battery is the potential difference
between its two terminals
+ -
V
Voltage and Electric field
The energy given to a charge by a voltage is:
∆U =
qV
WAB =-∆U
F=qE (E is the electric field)
V = - Ed
}
}
Since WAB = Fd, WAB = (qE)d = -∆U = -q V
(if the field E is constant)
Units:
V (volts)
E (NC-1)
d (metres)
Electrical Field (E)
V
E= −
d
F
E=
q
Volts per metre
Newtons per Coulomb
units are equivalent
Volts
Newton
=
metre Coulomb
V=
U
q
Newton
 Joule  1
=


 Coulomb  metre Coulomb
Joule
= Newton
metre
=
Joule Newton × metre
Electric field
Example :
The potential at the ground is zero. A storm
cloud has a potential of -50kV at an altitude
of 500m.
What is the associated electric field ?
-50kV
V = -Ed
500m
0V
E = -V/d = -(-50x103)/500 V/m
E = 100 V/m
Electric field
Example:
In the previous example, calculate the
speed of an electron reaching the ground.
Electron mass: 9.11x10-31kg.
The potential energy: qV
is transformed into kinetic
energy: ½mv²
1/2mv² = qV
2qV
v=
m
2(−1.6 × 10 −19 )(−50 ×103 ) −1
v=
ms
−31
9.11×10
v = 1.3 ×10 ms
8
−1
For the expert: v~40% of speed of light.
Relativistic effects become important!
Electrical Capacitance
+
+
+
+
+
+
V
V
A pair of metal plates separated by an
insulator. When subjected to a potential V,
charges ±Q will accumulate on the two plates.
Electrical capacitance (denoted C) is the
ability to store charge, expressed as ratio of
charge to potential difference:
Q
where Q is the charge on either plate C = V
SI unit of capacitance is coulomb per volt:
given the name farad (F) (Michael Faraday
1791-1867.- English physicist & chemist)
Voltage removed: charge remains on the plates
Electrical Capacitance
Parallel Plate capacitor
A uniform electric field
is created between
the plates E=V/d
Q
C =
V
Parallel plates separated by
free space (≈ air)
capacitance is given by
d
+
+
+
+
+
+
V
Aε 0
C=
d
- A is the common surface area of the plates
- d is their separation
- εo= 8.85 x 10-12 C2N-1m-2
is the “permittivity of free space”
Electrical Capacitance
Parallel Plate capacitor
+
+
+
+
+
+
+
+
+
+
+
+
-
-
Parallel plates separated
by free space (≈ air)
Aε 0
C=
d
To increase the capacitance
insert insulating material
between plates
Material referred to
as a dielectric
k known as the dielectric constant
Substance
Dielectric Constant
Vacuum
1.0
Air
1.00059
paper
3.7
glass
5.6
kAε 0
C=
d
Example
Calculate the capacitance of a parallel plate
capacitor of area A =5 cm2 if the plates are
separated by a material of thickness d= 0.1 cm
and dielectric constant k = 4? If the capacitor is
connected to a 9 volt battery, what is the resulting
charge on the positive plate? ε0 = 8.85x10-12C2N-1m-2
kAε 0
C=
d
C = 4 ∗ 5x10-4m2 ∗ 8.85x10-12 C2N-1m-2
0.1 x10-2m
C = 17.6x10-12F = 17.6pF
C = Q/V
Q = CV
Q = 17.6 x10-12F *9V = 1.58x10-10 Coulombs
Electrical Capacitance
Electrical energy (U) stored in a capacitor
+ + + Q
+ V=
ave
+ + C
U = QV
V
Voltage Applied:
Capacitor charges from zero to voltage V,
(V − 0 ) = 1 V
average voltage (Vave) during charging = 2
U = 1 QV
2
2


Q
1
U=
2 C 


2
U = 1 CV 2
2
Applications
•Defibrillator
•Photographic flash unit
Electrical Capacitance
+
+
+
+
+
+
V
The plates are moved apart while the battery
remains connected.
What happens to the capacitance C?
C decreases.
Aε 0
C=
d
What happens to the potential difference?
V is constant (battery remains connected).
What happens to the charge?
Charge Q decreases.
Q = CV
Electrical Capacitance
+
+
+
+
+
+
V
What happens to the charge if battery is
disconnected? Charge Q remains constant
Now the plates are moved apart while the
battery remains disconnected.
What happens to the charge?
Charge Q remains constant.
What happens to the capacitance C?
Aε 0
C=
d
C decreases.
What happens to the potential difference?
V increases.
Q
V=
C
Electrical Capacitance
Applications:
Computer keyboard
kAε 0
C=
d
Key
d
S
plates
Push key
•moves plates closer together
•Capacitance changes (increases)
•detected by computer electronics
Random access memory (RAM): Capacitors
used in RAM chips to store bits.
Touchscreen. A grid of small microscopic
capacitors to sense touch.
Electrical Capacitance
Electric energy stored in a capacitor
Application
Defibrillator
Ventricular fibrillation
Fast uncoordinated twitching of the heart muscles
Remedy
•Strong jolt of electrical energy
•Restores regular beating of the heart
Defibrillator
•Electrical energy stored in a capacitor
•Energy released in ≈ few milliseconds
Electrical Capacitance
Typical Defibrillator (AED)
150µF capacitor charged to 2250V
Magnitude of charge on each plate
−6
Q=
CV =
(150 ×10 F )(2250V ) =
0.3375C
Electric energy stored in capacitor
U = 1 CV 2
2
2
1
=
U
150 µ F )(×2250V=
) 380 J
2(
≈200J passed through body in ≈2ms
Therefore power in electrical pulse≈ 100kW
Defibrillator
•Electrical energy stored in a capacitor
•Energy released in ≈ few milliseconds
Electrical Capacitance
Application
Typical Photographic flash unit
Capacitor (300µF) charged to voltage of 300V
Electric energy stored
U = 1 CV 2
2
2
U = 1 300 µ F × ( 300V ) = 13.5 J
2

Energy stored is released rapidly ≈ 10-3 s
Power output of flash unit is ≈13.5 kW
Electric Current
The electric current (denoted I) is the charge
flowing per second:
Q
I=
t
It is a measure of the flow rate of charges
(analogous to the flow of liquid).
SI unit of current is ampere (I) coulomb/second
Named after French physicist André Ampere.
Note:
In electrical cables,
charges flow to an
charged capacitor,
appliance when switched
charges don’t move,
on:
thus I = 0 amps
current I = 0
+
+
+
+
+
+
-
electrons
Electric Current
Electric current not just confined to cables
Lightening strike – large electric field causes air
to ionize- large current of very short duration
At high electric fields
insulators may
become conducting:
dielelectric
breakdown
high energy electrons
This image cannot currently be displayed.
x rays
+
High dc voltage ≈50kV
Example
A defibrillator passes 50 A through the heart
for a period of 0.002 s. What is (a) is the quantity
of charge passed through the heart in this period?
(b) the capacitance of the capacitor supplying
this current if it operated at a potential difference
of 2000V?
q
Current =charge /time I =
t
Charge q = 50 A x 0.002s = 0.1C
Capacitance =Q/V = 0.1C/2000V= 50µF
Electric Current
Example
Current in a torch bulb is 0.2A.
How many electrons flow through the bulb
if it is on for 5 minutes?
Current =Charge/time
I = Q/t
0.2A = Q/(5*60)
Q = 5*60*0.2 = 60 Coulombs
But charge on the electron =1.6 x10-19C
Therefore number of electrons
= 60/(1.6 x10-19)
= 37.5x1019 electrons
7. The cell membrane in a nerve cell can be approximated
by a parallel plate capacitor with a surface charge density
of 5.9 x10-6 Cm-2. Determine the electric field within the
membrane. ε0 = 8.85x10-12 C2N-1m -2
Aε 0
C=
d
Q Aε 0
=
V
d
C=
Q
V
V
E= −
d
V
Q = Aε 0 = − Aε 0 E
d
Q 1
E = − 
 A  ε0
E
=
−6
−2
5.9
×
10
Cm
(
)
5
−1
=
6.6
×
10
NC
8.85 ×10−12 C2 N -1m -2
Example
It requires 5 Joules of energy to move a
positive charge of 0.1C from point A to point B.
Determine the potential difference between
A and B?
Q
+
q1
+
B
q2
+
A
energy
Potential difference V = charge
V = (5 J)/0.1C = 50 Volts