General Physics 2 Week 3 4
General Physics 2 Week 3 4
General Physics 2 Week 3 4
GENERAL PHYSICS 2
SUPPLEMENTARY SELF-LEARNING
MATERIAL
Semester 2
Week 3-4
REFERENCES:
Websites:
https://www.bbc.co.uk/bitesize/guides/z2b9hv4/revision/2
https://nustem.uk/activity/
https://www.physicsclassroom.com/
Books
Beiser, A. Fundamentals of Modern Physics. Fourth Edition
Resnick, R. H. (n.d.). Physics, 5th Edition. USA.
Halliday, D., Resnick, R. at al., Fundamentals of Physics Extended. 6th
Edition. Wiley. USA
Serway, R., Hewett,J. Physics for Scientists and Engineers with Modern
Physics.6th Edition.USA
Simpson, D. (2019, September 11). General Physics I: Classical
Mechanics. Largo, M
Recommended videos: I f internet connection is available, you can
watch the topics discussed and further explanations in videos by Michel
van Biezen on: http:www.ilectureonline.com/lectures/subject/PHYSICS
( by Michel van Biezen)
Or you may w atch through Y outube by keying a keyw ord and M i-
chael van Biezen Ex. “Vectors Michael van Biezen”
INSTRUCTIONS
Concepts:
Electric potential (V) Equipotential Points
Electric potential is a property of an If the points in an electric field are all
electric field, regardless of whether a at the same electric potential, then
charged object has been placed in that they are known as the equipotential
field. points. If these points are connected
Equal to the potential energy per unit by a line or a curve, it is known as an
charge and is expressed in Joules per equipotential line. If such points lie on
Coulomb (J/C) or volt. a surface, it is called an equipotential
V= U/q surface.
Electric Potential energy (U) Relationship Between Electric
The energy of charged object in an Potential and Electric Field
external electric field ( or more
ΔV=-EL
precisely, the energy of the system
consisting of the object and the Electric field vs Electric
external electric field); measured in Potential
Joules (J) or in electron volt (eV)
where 1eV=1.60 x10-19 J The electric field, E, is a measure
of force per unit charge ( E=F/q);
Equipotential surfaces the electric potential ,V, is a
Any surface with the same electric measure of energy per unit
potential at every point is termed as
an equipotential surface.
charge (V=U/q)
3
W= qEL cos θ
where;
At which point (A or B) is the potential
W-work done energy larger,
q– magnitude of the charge a) For a positive charge +q ?
d or L– displacement of the charge b) For a negative charge –q ?
θ– angle between E and L
4
Equipotential lines
In this illustration, the
Watch more of this on YouTube: electric field lines are shown emanating
from a positive charges ( right plate) and
Physics - Electrical Potential and Electrical terminating in the negative charges ( left
Potential Energy by Michel van Biezen plate) . Electric field lines are also shown
Part 1 is here: starting from high potential and ending in a
low potential. Dashed lines shown
https://www.youtube.com/watch? perpendicular to the electric field lines are
v=eZPmIKneSBk&t=16s equipotential lines. Work done along these
lines is zero.
Let : F=qE
W=Fd cos θ
W=qEd cos θ; if E and d are
perpendicular, then θ=900
W=Fd cos 900 =0
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Sample problem:
1. A particle with a charge , q=3.0x10-9 C 2. Dry air will support a maximum
moves from point a to point b along a electric field strength of about
straight line, a total distance, d= 0.50m. 3.0 × 106 V/m. Above that value, the
The magnitude of the electric field, E=200 field creates enough ionization in the
N/C. Determine the (a) force on q (b) work air to make the air a conductor. This
done on it by the field and (c) the potential allows a discharge or spark that
difference Va-Vb . reduces the field. What, then, is the
maximum voltage, Vab, between two
parallel conducting plates separated
E=200 N/C by 2.5 cm of dry air?
Given: E=3.0 × 106 V/m
q=3x10-9 C
d=2.5 cm =0.025 m
Solution:
3. A solid conducting sphere with radius,
A.) The force is in the same direction as the
R, has a total charge Q, find the
electric field and its magnitude is given by:
potential, both outside and inside the
F=qE sphere.
F=(3.0 x10-9 C ) ( 200 N/C ) Solution: Let r-radius inside
-7 sphere and R –radius outside
F= 6.0 x10 N
sphere
Concepts:
Capacitors Charging a Capacitor
An electronic device that can store When a capacitor is charged, its plates
energy as a potential energy in an have equal but opposite charges of +
electric field. Q and –Q.
Spherical capacitors
A spherical capacitor consists of two
concentric conducting spheres of radii,
R1 and R2 .
Cylindrical Capacitors
A cylindrical capacitor consists of two
concentric, conducting cylinders of
length, L.
7
Capacitors
A capacitor is a device used to store
increases with the magnitude
electric charge. Capacitors have (amount) of charge and decreases
applications ranging from filtering static out with distance between charges. Thus,
of radio reception to energy storage in the bigger the plates are, the more
heart defibrillators. Typically, commercial
charge they can store—because the
capacitors have two conducting parts close charges can spread out more.
to one another, but not touching. When Thus, C should be greater for larger
battery terminals are connected to an area, A. Similarly, the closer the
initially uncharged capacitor, equal amounts plates are together, the greater the
of positive and negative charge, +Q and attraction of the opposite charges on
–Q, are separated into its two plates. The them. So, C should be greater for
capacitor remains neutral overall, but we smaller distance, d. It is clear then
refer to it as storing a charge Q in this that for a parallel- plate capacitor
circumstance. there are only two factors (A and d)
The amount of charge Q a capacitor can that affect its capacitance C. The
store depends on two major factors—the capacitance of a parallel plate
voltage applied and the capacitor’s physical capacitor in equation form is given by
characteristics, such as its size. And we can
say that, C=ε0A/d
Q=CV Where: ε0– permittivity of free space
Where: Q-amount of charge =8.85 × 10−12 F/m
C-capacitance of the capacitor A– area of the plates
V– voltage applied d– distance between plates
Parallel– Plate Capacitors The Farad (F) is the unit of
capacitance. Other smaller multiples
Two parallel metal plates (conductors)
are also used such as: nano Farad
isolated from each other and from their
(nF), micro Farad (μF) and the millli
surroundings form a capacitor. The
Farad (mF).
parallel-plate capacitor is the simplest type.
Sample Problem:
(a) What is the capacitance of a
parallel plate capacitor with metal
plates, each of area 1.00 m2,
separated by 1.00 mm?
(b) What charge is stored in this
capacitor if a voltage of 3.00
×103 V is applied to it?
Let:
The inner cylinder radius be R1
and the outer cylinder radius be R2
The capacitance of a cylindrical capacitor is
given by
C= 2∏ε0 L/ ln ( R2/R1)
And the electric field, E is given by
E=q/ 2∏ε0 L R
Where R is the radius of the Gaussian
surface.
Figure 1. (a) Capacitors connected in series.
The magnitude of the charge on each plate
A spherical capacitor is formed by two is Q. Connected one after the other in a
concentric conducting spheres with radii R1 single path, it is equivalent to a capacitor
and R2 . having a larger plate separation, d. Series
connections produce a total capacitance
that is less than that of any of the individu-
al capacitors.
Now, let us define the charge, voltage and
capacitance for capacitors in series.
Charge: The charge is the same
anywhere in the connection.
QT = Q1 = Q2 =Q3 ... where Q1=C1V1 …
and QT= CTVT
Voltage: The total voltage is the sum
of individual voltages.
Let: VT = V1 + V2 + V3... where V1=Q1/C1…
The inner sphere radius be R1 and VT= QT/CT
and the outer sphere radius be R2 Capacitance: The total capacitance is
the inverse of the sum of the inverses of
The capacitance of a cylindrical capacitor is
each individual capacitances.
given by
1/CT = 1/C1+1/C2+1/C3…
C= 4∏ε0 (R1R2 )/ ( R2-R1)
where “…” indicates that the expression is
And the electric field, E is given by
valid for any number of capacitors
E=q/ 4∏ε0 L R2 connected in series. An expression of this
form always results in a total
Where R is the radius of the Gaussian
capacitance CT that is less than any of the
surface.
individual capacitances.
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Sample problem: (b) What is the total charge and the charge
on each capacitor?
(a) Find the equivalent capacitance of three
capacitors, 2.0 μF, 4.0μF and 6.0 μF con- QT = CTVT
nected in series with a 24-V battery. QT = (1.09 x 10-6 F) (24 V)= 2.62 x 10-6 C
Solution: QT = 26.2 μC = 26 μC = Q1 = Q2 =Q3
1/CT = 1/C1+1/C2+1/C3 (c) What is the voltage across each
1/CT = 1/2.0 μF+1/4.0 μF+1/6 .0μF capacitor?
1. Solve for the total charge, the charge 2. Solve for the potential energy
on and the potential difference stored inside the capacitor given the
across each capacitor in a network of geometry and the potential difference
capacitors across the capacitor
Concepts :
A table of equations summarizing
capacitance, charge and voltage in
series and parallel capacitance circuits
The potential energy, U,
Quantity Series Parallel of a charged capacitor
U=1/2 QV ; since Q=CV
Capacitance 1/CT = 1/ CT =
C1+1/C2+1/ C1+C2+C3… Then,
(C)
C3 ... U=1/2 (CV)V
Unit: Farad
(F) U=1/2 CV2
VT = V1 + V2 + VT = V1 = V2 =
V3… V3… Some basic circuit
Voltage (V) symbols
VT= QT/CT VT= QT/CT
Unit: volt (V) In making a schematic diagram
Q1=C1V1 …. Q1=C1V1 …. of electronic circuitry, the
following symbols are used:
Q T = Q1 = Q2 QT = Q1 + Q2
=Q3 ... +Q3 … wire
Charge (Q)
QT= CTVT QT= CTVT
Unit: cell
Coulomb (C) Q1=C1V1 … Q1=C1V1 …
capacitor
Note: Do not confuse the
Capacitance, C with the Coulomb (C).
Capacitance is a quantity and Coulomb Earth ( ground)
is a unit.
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Sample Problem:
Three capacitors, C1, C2, and C3 whose QT = Q1 + Q2 +Q3
values are indicated in the illustration after we have computed the charge in each
below are connected in parallel with a capacitor.
24-V battery.
CT = 12 μF=12 x 10-6 F
Or CT = 1.2 x 10-5 F
One way to check your answer is to
remember that since you are adding the
individual capacitances, your total capaci-
tance should be greater than the greatest
value in the given capacitances.
Find :
(b) The total charge can be solved
using (a) the total capacitance ( 6 μF)
QT= (1.2 x 10-5 F) (24 V) (c) the charge in capacitor 1 (96 μC)
QT= 2.88 x 10-4 C (d) the voltage , V2, and V3 in the 3μF
and 6μF capacitors ( 16 V and 8 V
QT= 2.9 x 10-4 C
respectively ( assume all digits in
This answer can be verified later using the answer to be significant)
the equation
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Concepts:
Dielectric
An insulating material which when Dielectric and Electric field
placed in an electric field significantly
reduces the electric field than with Inserting an insulator between the
nothing ( vacuum) between them. plates reduces the net electric
field between the plates.
Dielectric constant, k
Dielectric breakdown
A measure of how effective a material
is in reducing an electric field set up A process that occurs when
across it. an electrical insulating material,
subjected to a high enough voltage,
Dielectric and Capacitance suddenly becomes an electrical
conductor and electric current flows
Inserting an insulator between the
through it. All insulating materials
plates reduces the electric field
undergo breakdown when the electric
flow within the gap separation. field caused by an applied voltage
This in turn reduces the voltage in exceeds the material's dielectric
the capacitor and increases the strength.
capacitor’s capacitance.
The table below, shows some materials and Inserting an insulator between the
their corresponding dielectric constant, K. capacitor plates reduces the electric
field flow within the gap separation.
This in turn reduces the voltage , V,
in the capacitor and increases the
capacitor’s capacitance.
Since; V=Ed, and C=Q/V; then
Q/C=Ed
This equation shows the inverse
relationship between electric field, E,
between the plates and the
capacitance, C, of the capacitor.
The voltage, V can also be computed
using these relationships. Since V and
C are inversely proportional, then
from
K=C/Co we can also deduce
that
K =Vo/V
where
Vo-voltage with no dielectric
V– voltage with a dielectric
Sample problem:
The parallel plates in a capacitor have
If we let Co to be the capacitance of the
an area of 2000 cm2 (2.00 x 10-1 m2 )
capacitor with no dielectric where
and are 1.00 cm (1.00 x10-2 m)
Co=ε0 A/d apart. The original potential difference
Vo between the plates is 3000 V and it
and C is the capacitance of the capacitor
decreases to 1000 V when a sheet of
with a dielectric having a dielectric
a dielectric is inserted between the
constant, k, where
plates. (a) What is the original
C=Kε0 A/d capacitance Co of the capacitor ? (b)
What is the dielectric constant of the
Then, we can also say that
dielectric material used?
C=KCo ; K=C/Co
Given:
Which means that that capacitance of the
Plate area = 2000 cm2
capacitor with a dielectric is increased K
times its capacitance with no dielectric. Distance between plates=1.00 cm
For example, say, a parallel-plate capacitor Vo =3000 V
with no dielectric has a capacitance Co=10
V=1000 V
nF. If paper, having a dielectric constant,
K=3.5 is inserted between the plates of this Solution:
conductor its new capacitance value will
(a) Co= εo A/d
increase 3.5 times or we can say that
Co= 8.85 x10 –12 F/m (2.00 x
C=(3.5) ( 10 nF)
10-1 m2 ) /1.00 x10-2 m
C= 35 nF
Co=1.77 x10-10 F
For the same capacitor area and distance of
(b) K= Vo/V
separation between plates, a capacitor with
a dielectric can have 3.5 times higher K= 3000 V/ 1000V
capacitance!
K= 3 – From this answer
we can also guess that the capacitance of
the capacitor increased 3 times its Co .
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WEEK 3 DAY 4
(60 mins)
A. Electric Potential
1. What is the electric potential at B. Capacitors
Point P, located at the center of 1. How large should a capacitor be
the square of point charges. The for it to have a capacitance of 1F
distance d= 1.3 m and the charges
are: if the distance between its plates
is 1mm?
q1=+12 nC q2=-24 nC
q3=+31 nC q4=+17 nC
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15
WEEK 4 DAY 4
(60 mins)