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phy

phy 102

Question	Answer
Electric charge
• Intrinsic property of the particles that make
up matter
Electric charge
• Charge can be positive or negative
Electric charge
• Atoms are composed of negatively-charged
electrons and positively-charged protons
Electric charge
• Charge is measured in Coulombs [unit: C]
Electric charge
• Charge is measured in Coulombs [unit: C]
• Proton and electron have equal and
opposite elementary charge = 1.6 x 10-19 C
• Charge on proton = +1.6 x 10-19 C
• Charge on electron = -1.6 x 10-19 C
Electric charge
• We now know that protons and neutrons
are made up of quarks with 2/3 and -1/3
charges (electrons are still fundamental)
Electric charge
• Charge cannot be created or destroyed (it is
conserved) but it can be moved around
A balloon is rubbed against a nylon
jumper, and it is then found to cause a
force of attraction to human hair.
From this experiment it can be
determined that the electrostatic
charge on the balloon is
1. positive
2. negative
3. Impossible to determine
0%
0%
0%
1.
2.
3.
Electric charge
• Charges feel electrostatic forces
Electric charge
• Rub a balloon on your hair and it will stick to things! Why??
• Friction moves electrons from your hair to the balloon
• The balloon therefore becomes negatively charged, so your
hair becomes positively charged (charge conservation)
• Your hair will stand on end (like charges repel), and the
balloon will stick to your hair (opposite charges attract)
• Now move the balloon near a wall. The wall’s electrons are
repelled, so the wall becomes positively charged.
• The balloon will stick to the wall! (opposite charges attract)
Electric charge
• Rub a balloon on your hair and it will stick to things! Why??
Electrostatic force
• The strength of the electrostatic force between
two charges q1 and q2 is given by Coulomb’s law
𝐹𝑒 𝐹𝑒
𝑘 =9×109𝑁𝑚2𝐶−2
• The direction of the force is along the joining line
Electrostatic force
• The electrostatic force is a vector, written Ԧ
𝐹
• Vectors have a magnitude and a direction. This
may be indicated by components Ԧ 𝐹 = (𝐹𝑥,𝐹𝑦,𝐹𝑧)
Ԧ
• The magnitude is sometimes written as
can be evaluated as \| Ԧ 𝐹\| =
𝐹 . It
�
�𝑥
2 +𝐹𝑦
2 +𝐹𝑧
2
• The direction can be indicated by a unit vector
Electrostatic force
Example
Two 0.5 kg spheres are placed 25 cm apart. Each sphere has a
charge of 100 μC, one of them positive and the other negative.
Calculate the electrostatic force between them and compare it to
their weight.
𝐹=𝑘\|𝑞1\|\|𝑞2\|
𝑟2 𝑘=9×109𝑁𝑚2𝐶−2 Coulomb’s Law:
\|𝑞1\|=\|𝑞2\|=100𝜇𝐶=100×10−6𝐶=10−4𝐶
𝑟=25𝑐𝑚=0.25𝑚
𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐=9×109×10−4×10−4
0.252 =1440𝑁
𝐹𝑤𝑒𝑖𝑔ℎ𝑡=𝑚𝑔=0.5×9.8=4.9𝑁
Electrostatic force
• Where multiple charges are present, the forces
sum as vectors (“principle of superposition”)
+ve
+ve
+ve
What is the combined force
on the blue charge from the
two red charges?
Electrostatic force
• Where multiple charges are present, the forces
sum as vectors (“principle of superposition”)
+ve
+ve
+ve
𝐹2
𝐹1
𝐹𝑡𝑜𝑡𝑎𝑙 =
𝐹1 +
𝐹2
Electrostatic force
• Where multiple charges are present, the forces
sum as vectors (“principle of superposition”)
+ve
+ve
+ve
𝐹2
𝐹1
Electric field
• The electric field at a point is the force a unit
charge (q = +1 C) would experience if placed there
𝐸 =
Ԧ 𝐹
𝑞 Ԧ 𝐹=𝑞
𝐸
(Units of E are N/C)
• It is a vector and its direction can be represented
by electric field lines
• Let’s look at some simple examples!
Electric field
• Electric field around a positive charge +Q
+q
Test charge +q at separation r
feels an outward force
\|𝐹\| =
𝑘 𝑄𝑞
𝑟2
Electric field is also outward
\|𝐸\| =
\|𝐹\|
𝑞
=
𝑘 𝑄
𝑟2
Now imagine placing the test charge at many different
places to map out the whole electric field
Electric field
• Electric field around a positive charge +Q
Magnitude of electric field at
any point:
\|𝐸\| =
\|𝐹\|
𝑞
=
𝑘 𝑄
𝑟2
Direction of electric field is
radially outward
Electric field
• Electric field around a negative charge -Q
Magnitude of electric field at
any point:
\|𝐸\| =
\|𝐹\|
𝑞
=
𝑘 𝑄
𝑟2
Direction of electric field is
radially inward
Electric field
• Electric field lines start on positive charges and end
on negative charges
• The more closely spaced the field lines, the
stronger the force
Electric field
• The direction of the field lines show how a positive
charge would move if placed at that point. A
negative charge would move the opposite way.
𝐸
+q
Ԧ
𝐹 =−
𝐸/𝑞-q
Ԧ
𝐹 =
𝐸/𝑞
Electric field
• Electric field lines between two charges
Unlike charges
Like charges
Electric field
• Electric field lines between charged plates
Electric field
• Electric field lines between charged plates
𝐸
• A constant electric field is obtained (see later
material on capacitors)
Electric field
ExampleA +5.0 mCcharge is located at the origin,
and a -2.0 mCcharge is 0.74 m away on the x-axis.
Calculate the electric field at point P, on the y-axis
0.6 m above the positive charge. If a +1.5 mCwas
placed at P, what force would it experience?
0.74 0
0.6
P
Electric field at P due to green charge q = +5x10-6C
Electric field is superposition of 2 charges
0
P
0.6
𝐸=𝑘𝑞
𝑟2=9×109×5×10−6
0.62 =1.25×105𝑁/𝐶
Direction is along y-axis: 𝐸𝑥,𝐸𝑦 =(0,1.25×105)
E= kq/r2along joining line, k=9x109
Electric field
Example
A +5.0 mC charge is located at the origin,
and a -2.0 mC charge is 0.74 m away on the x-axis.
Calculate the electric field at point P, on the y-axis
0.6 m above the positive charge. If a +1.5 mC was
placed at P, what force would it experience?
Electric field is superposition of 2 charges
E= kq/r2 along joining line, k=9x109
0.6
P
0.6
P
0
0.74
Electric field at P due to purple charge q = -2x10-6 C
�
�=
𝑘 \|𝑞\|
𝑟2 Pythagoras: r2 = 0.62 + 0.742 = 0.91 m2
0.74
r = 0.95 m
Electric field
ExampleA +5.0 mCcharge is located at the origin,
and a -2.0 mCcharge is 0.74 m away on the x-axis.
Calculate the electric field at point P, on the y-axis
0.6 m above the positive charge. If a +1.5 mCwas
placed at P, what force would it experience?
0.74 0
0.6
P
0.74
P
Electric field is superposition of 2 charges
E= kq/r2along joining line, k=9x109
Electric field at P due to purple charge q = -2x10-6C
0.6
𝐸=𝑘\|𝑞\|
𝑟2 =9×109×2×10−6
0.952 =0.20×105𝑁/𝐶
Electric field
ExampleA +5.0 mC charge is located at the origin,
and a -2.0 mC charge is 0.74 m away on the x-axis.
Calculate the electric field at point P, on the y-axis
0.6 m above the positive charge. If a +1.5 mC was
placed at P, what force would it experience?
0.74 0
0.6
P
Electric field is superposition of 2 charges
E= kq/r2along joining line, k=9x109
Electric field at P due to purple charge q = -2x10-6C
�
�=𝑘\|𝑞\|
𝑟2 =9×109×2×10−6
0.952 =0.20×105𝑁/𝐶
0.74
0.6
𝐸𝑥,𝐸𝑦 =(0.16×105,−0.13×105)
Electric field
Example
A +5.0 mC charge is located at the origin,
and a -2.0 mC charge is 0.74 m away on the x-axis.
Calculate the electric field at point P, on the y-axis
0.6 m above the positive charge. If a +1.5 mC was
placed at P, what force would it experience?
Electric field is superposition of 2 charges
Green charge:
Purple charge:
�
�𝑥,𝐸𝑦 = (0,1.25×105)
0.6
P
0
𝐸𝑥,𝐸𝑦 = (0.16×105,−0.13×105)
Total:
𝐸𝑥,𝐸𝑦 = (0.16×105,1.12×105)
Electric field strength at P:
𝐸 =
0.74
𝐸𝑥
2 +𝐸𝑦
2 = 1.13×105 𝑁/𝐶
𝐹 =𝑞𝐸 =1.5×10−6×1.13×105 =0.51𝑁
Force:
Electric dipole
Dipole moment
Electric dipole
• A dipole in an electric field will feel a torque
but no net force
𝐸
𝜏 =𝐹𝑙sin𝜃 = 𝐸𝑄𝑙sin𝜃 Ԧ 𝜏=
𝐸 × Ԧ 𝑝
Electrostatic
analyzer
• Charged particles will experience a force in an
electric field F=qE, hence acceleration a=F/m=qE/m
Electrostatic analyzer
•An electrostatic analyzerselects velocities
Uniform electric field E applied
between curved surfaces
r
𝑎= 𝐹
𝑚=𝑞𝐸
𝑚
𝑎=𝑣2
𝑟
Acceleration a is given by:
𝑣2
𝑟 =𝑞𝐸
𝑚→𝑣= 𝑞𝐸𝑟
𝑚
Conductors and Insulators
• In metals (e.g. copper, iron) some electrons are weakly held
and can move freely through the metal, creating an electric
current. Metals are good conductors of electricity.
Conductors and Insulators
• In metals (e.g. copper, iron) some electrons are weakly held
and can move freely through the metal, creating an electric
current. Metals are good conductors of electricity.
• In non-metals (e.g. glass, rubber, plastic) electrons are
strongly held and are not free to move. Non-metals are
poor conductors of electricity, or insulators.
• Semi-conductors (e.g. germanium, silicon) are half-way
between conductors and insulators.
Freely moving electrons make metals good conductors of electricityand heat
Summary
• Matter is made up of positive and negative charges.
Electrons/protons carry the elementary charge 1.6 x 10-19 C
𝐹=
• Forces between charges are described by Coulomb’s Law
𝑘 𝑞1 𝑞2
𝑟2 𝑘=9× 109𝑁𝑚2𝐶−2
• Forces from multiple charges sum as vectors
• Electric field describes the force-field around charges
𝐸 =
Ԧ 𝐹
𝑞 Ԧ 𝐹=𝑞
𝐸
1. Two 0.5 kg spheres are placed few meters apart. Each sphere
has a charge of 60 μC, one of them positive and the other
negative. If the electrostatic force between them is 1400 N,
calculate the distance between them. 5 marks
2. You measure an electric field of 1.25 X 106 N/C at a distance
of 0.150 m from a point charge. There is no other source of
electric field in the region other than this point charge. (a)
What is the electric flux through the surface of a sphere that
has this charge at its center and that has radius 0.150 m? (b)
What is the magnitude of this charge? 6 marks
3. A parallel-plate air capacitor is to store charge of
magnitude 240.0 pC on each plate when the potential
difference between the plates is 42.0 V. (a) If the area of
each plate is 6.80 cm2, what is the separation between the
plates? (b) If the separation between the two plates is
double the value calculated in part (a), what potential
difference is required for the capacitor to store charge of
magnitude 240.0 pC on each plate? 6 marks
4. What is the resistance of a carbon rod at 25.8°C if its
resistance is 0.0160 Ω at 0.0°C? 3 marks (Take α for
carbon as– 0.0005 [(°C)−1])
Lectures I and II:
Electric charge, force, field
• What is electric charge and how do we
measure it?
• Coulomb’s Force Law between charges
• How an electric field can be used to describe
electrostatic forces
• Some simple applications of these principles

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