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Lecture 5
o Aim of the lecture
 More detail on
 Capacitance
 Ohms Law
 Capacitance
 Energy storage
 Dielectrics
 Ohms Law
 resistance
 Power dissipation
o Main learning outcomes
 familiarity with
 Dielectrics
 Resistance
 Addition of
Capacitors
 Resistors
 Power dissipation in resistors
Dielectrics
Material between plates
oA dielectric is
o an insulator
o either polar
o or non-polar
No dielectric between plates C = e0A/d
With a dielectric, then C = ere0A/d
+
-
Polar Dielectric
Electrons leave one plate
The same number arrive on the other
+ve
o In a polar material
 molecules are polarised
 act like small dipoles
 orientate to align with E field
E
-ve
[pure water is like this]
Polar Dielectric
Consider Gauss’ Law
surface as shown
net charge inside surface
reduced
+ve
E
charge = e0
-ve
q-q’ = e0
∫ E.dA
But charge is now
q-q’ where
q is the charge without dielectric
q’ is the charge due to polar molecules
∫ E.dA
Polar Dielectric
+ve
q-q’ = e0
∫ E.dA
E
-ve
which gives that the electric field
is E = (q-q’)/e0A
compared with
E0 = q/e0A
with no field
The electric field is weaker when a dielectric is present for the same
applied voltage
The quantity of charge q’ depends on E0, (often) which depends on q
so q’ is proportional to q
Polar Dielectric
+ve
so q’ = const q
and
E
-ve
E = const E0
define that 1/const is er
with the result that C = ere0A/d
The quantity of charge q’ depends on E0, (often) which depends on q
so q’ is proportional to q
Other geometries are possible
cylindrical
Sphere above a plane
A non-polar dielectric is one where the molecules
are non-polar
o In this case the molecules are CAUSED to be polar by
the electric field, they are
 Induced dipoles
In practice the only difference is that the values of er
are (usually) smaller than for a polar dielectric
Adding Capacitance
The net
charge on these
two plates is
zero, it is just
an equipotential
line
Add a second capacitor in series
Adding Capacitance
Overall effect is to
double distance between plates
but C = ere0A/d
so capacitance is halved
C
Total capacitance = C/2
C
Adding capacitors in series
REDUCES the total capacitance
More generally
When adding capacitors in series
1
CT
=
1
C1
+
1
C2
Example: capacitance with partial dielectric
Capacitors Connected in Parallel
o Adding a secondbut
identical
in parallel
C = erecapacitor
0A/d
 doubles the area of the plates
 doubles charge
stored for same
voltage applied
so capacitance
is doubled
More generally
When adding capacitors in parallel
CT =
C1 +
C2
CT =
1
CT
=
C1 +
1
C1
+
C2
1
C2
PARALEL
SERIES
Energy Stored
The energy stored is equal to the energy in the
electric field between the plates.
Q = CV
Work done to move a small charge, dq
from one plate to the other is
dW = VdQ = VCdV
So total energy, E is
E = ∫dW = C∫VdV = ½CV2
Summary for Capacitors
Q = CV
E = ½CV2
CT =
C1 +
C2
1
1
1
CT
=
C1
C = ere0A/d
+
C2
For parallel addition
For series addition
for parallel plate capacitor
e0 is 8.854×10−12 F m–1
er is typically between 1 and 10
Practical Information:
Capacitors are labelled
in a ‘funny’ way
TheMore
units than
are always
you
mF wanted
or pF to know!
And this means the maximum
voltage you can put across
it is 100V
examinable
This not
means
100k pF
ie 100 x 103 x 10-12 F = 100nF
but useful
(it cant be mF because it would
be too big – you just have to
know this!)
Final Warning:
some capacitors are ‘polarised’
you MUST put the voltage across them
the correct direction.
This little + means
that this terminal
must be +ve compared
with the other
Tantalum bead capacitors
can explode if connected
the wrong way round!!
22u
35
means 22mF
max volts = 35V
More on Ohms Law and Resistance
Recall that
V = IR
Where V is the voltage applied across a resistance, R
and I is the current that flows.
The resistance is analogous to the resistance of a pipe
to the flow of water through it.
o Electrons are made to drift in an electric field caused by an external voltage.
o They loose energy in collisions with the fixed atoms
o They therefore do not accelerate
o They drift at constant speed
Consider a resistance with a voltage across it.
V
Suppose the current that flows is Ia
Current = 2Ia
If we apply the same voltage across two such resistances
connected in parallel, then the current doubles, so
the resistance is inversely proportional to the area, A, of conductor
If we put two in series, then we need a voltage
V across each to drive the same current, Ia, so
resistance is proportional to length, L
Resistance
= r L/A for many materials
r is a constant called the ‘resistivity’ of the material
and is very different for different conductors.
V=IR
o Any real circuit has resistance
o Usually wires are a small resistance
 we ignore it, assume it is zero
 Represent the resistance with a
 RESISTOR
o The wires that we draw joining parts of a circuit are
 Taken to have zero resistance
 Resistance is represented by
o An ampmeter has zero resistance
o A voltmeter has a very high resistance (infinite if perfect)
More practical details!
Here is how you can tell what a resistor value is:
These resistors are
100W with 5% accuracy
Resistors in series.
Recall that for identical resistors, the resistance is proportional
to the length. This generalises for resistors in series
Rtotal = R1 + R2 + R3 + …..
For resistors in parallel, it is like increasing the area,
so two in parallel gives half the resistance, and in general:
Rtotal
The energy transferred to the atoms when the
electrons collide with them in a resistor is converted to heat
Power = current x voltage
P = IR
but remember ohms law V=IR
So
P = I2R
P = V2/R