Download MORE ELECTROSTATICS Coulomb`s Law Gives the force that two

MORE ELECTROSTATICS
Coulomb’s Law
Gives the force that two electric charged particles exert on each other
Formula is
F=
kq1q2
r2
F = the electrostatic force between the charges (Newtons)
q1 = charge on particle 1 (Coulombs)
q2 = charge on particle 2 (Coulombs)
r = distance between particle 1 and 2 (meters)
k = a constant = 8.99 x 109 Newton-meters2/Coulomb2
If F > 0, force is repulsive. If F < 0, force is attractive.
Example. How much force does a spec of dust whose net charge is 2 x 10-5 Coulombs exert
on a second spec of dust whose net charge is -3 x 10-5 Coulombs and is 0.5 meters away from
the first spec?
q1 = 2 x 10-5 Coulombs
q2 = -3 x 10-5 Coulombs
r = 0.5 meters
F=
(8.99 × 109 Nm 2 / C 2 ) × (2 × 10−5 C ) × (−3 × 10−5 C )
= -21.6 N
(0.5m) 2
The minus sign indicates that it is an attractive force.
Q. How much force does the second spec exert on the first spec?
A. Exactly the same as the first exerts on the second, 21.6 N [Of course, it is in the opposite
direction.]
Kinetic and Potential Energy
Kinetic Energy
Energy due to motion
Depends on velocity (of the object possessing the energy)
Formula:
EK = 12 mv 2
EK = kinetic energy (Joules)
m = mass of object that possesses the kinetic energy (kg)
v = velocity of the object (m/s)
Formula only holds for velocities much smaller than the speed of light
Example. A pickup truck weighs 2 metric tons and travels at 60 kilometers per hour.
What is its kinetic energy?
m = 2 metric tons = 2 x 1000 kg = 2000 kg
v = 60 kilometers / hour = 60 x 103 m / (3600 sec)
= 16.7 m/s
EK = 0.5 x (2000 kg) x (16.7 x m/s)2
= 2.79 x 105 kg-m2/s2 = 2.79 x 105 Joules
Potential Energy
Energy that has the potential to become kinetic energy.
Depends on the position (of the object possessing the energy).
Example 1: A rock in a gravitational field.
You hold a rock in your hand. It has potential energy, because if you
release it, it starts to move. That is, its potential energy becomes kinetic
energy. The farther up from the ground you hold the rock, the more
potential energy it has. [The higher you hold it, the faster it will be going
when it hits the ground, i.e. the more kinetic energy it will have.]
Example 2: An electron an electric field.
An electron is 0.5 m away from positively charged metallic plate (an
anode) inside a huge vacuum tube. The electron is pulled toward
the anode. That is, it has potential energy that is being converted into
kinetic energy. The farther the electron is from the anode, the more
potential energy it has. [The farther away it is, the faster it will be going
when it it’s the anode, i.e. the more kinetic energy it will have.]
Example 3: Potential Energy of a Charged Particle Due to Another Charged Particle
Φ=
kq1q2
r
Φ = potential energy (Joules)
q1 = charge of first particle
q2 = charge of second particle
r = distance between particles.
k = 8.99 x 109 Newton-meters2/Coulomb2
Q. An electron within a vacuum tube is 0.25 meters away from the
positively charged anode. If the charge on the anode is 2 Coulombs,
what is the the potential energy of the electron?
A.
q1 = 2 C
q2 = 1.6 x 10-19 C (the charge of the electron)
Φ=
(8.99 ×109 Nm 2 / C 2 ) × (2C ) × (−1.6 ×10−19 C )
0.25m
= -1.15 x 10-8 Nm = -1.15 x 10-8 Joules
Total Energy
= (Kinetic Energy) + (Potential Energy)
In the absence of friction or other external forces, the total energy of a system is
constant.
Example.
An electron in a vacuum tube is 0.25 meters away from the
positively charged anode. The charge on the anode is 2 x 10-5
Coulombs. The electron is initially stationary. What will be the velocity of
the electron after it travels 0.01 meters towards the anode?
Strategy: Calculate the potential energy that electron has when it’s
a) 0.25 m away from the anode, and b) 0.24 meters away from it.
Equate the difference in these potential energies with the kinetic energy.
Calculate the velocity from the kinetic energy.
Principle used: Total Energy is constant. So the loss of potential energy
must be the gain in kinetic energy.
Initial Potential Energy
Φi =
(8.99 ×109 Nm 2 / C 2 ) × (2 ×10−5 C ) × (−1.6 ×10−19 C )
0.25m
= -1.15 x 10-13 Joules
Final Potential Energy (distance from anode = 0.25 m – 0.01 m = 0.24 m)
Φf =
(8.99 ×109 Nm 2 / C 2 ) × (2 ×10−5 C ) × (−1.6 ×10−19 C )
0.24m
= -1.20 x 10-13 Joules
Gain in Kinetic Energy = Loss of Potential Energy
1
2
mv 2 = Φ f − Φ i
(m = mass of electron)
Solve for v
v=
=
=
2
Φ f − Φi
m
2
(−1.20 × 10−13 J ) − (−1.15 × 10−13 J )
−31
9.11× 10 kg
2 × (0.05 × 10−13 J )
9.11× 10−31 kg
= 1.04 x 108 m/s
Electric Potential
Potential energy per charge.
Depends on position.
The potential field (position-dependent function) produced by a particular
charge depends only on that charge.
Formula for the potential field produced by a charge q
φ =
kq
r
(potential due to q at a distance r)
φ = potential (Joules/Coulomb = Volts)
q = charge (Coulomb)
r = distance from q (meter)
k = 8.99 x 109 Newton-meters2/Coulomb2
Example. What is the potential 3 meters away from a 2 Coulomb charge?
φ=
(8.99 ×109 Nm2 / C 2 ) × (2C )
3m
= 5.99 x 109 Nm/C = 5.99 x 109 J/C = 5.99 x 109 Volts
What is a the potential energy of an electron 3 m away from this charge?
Potential = (Potential Energy)/Charge.
So
(Potential Energy) = Potential x Charge
= (5.99 x 109 Volts) x (1.6 x 10-19 C)
= 9.58 x 10-10 Volts C = 9.58 x 10-10 Joules
Example. The difference between the potential at the anode (positively charge plate) of a
vacuum tube and the cathode (negatively charged plate) of the tube is 100 KiloVolts. What is
the kinetic energy of an electron when it strikes the anode after leaving the cathode and being
accelerated across the tube?
Strategy: Obtain the potential energy lost. This must equal the kinetic energy gained.
Potential energy lost = | (final potential energy) – (initial potential energy)|
= | (final potential) x charge – (initial potential) x charge |
= | (final potential) – (initial potential) | x charge
= | potential difference| x charge
= (100 KiloVolts) x (1.6 x 10-19 C)
= (100 x 103 Volts ) x 1.6 x 10-19 C
= 1.6 x 10-14 Volts C
= 1.6 x 10-14 Joules = kinetic energy gained
What is this energy in electron Volts?
1 eV = 1.6 x 10-19 Joules
1 Joule = 1/(1.6 x 10-19) eV = 6.25 × 1018 eV
So
1.6 x 10-14 Joules = 1.6 x 10-14 (6.25 x 1018 eV)
= 1.0 x 105 eV
Electric Field
Force per charge
Generally depends on position (but not always)
The electric field produced by a particular charge only depends on that charge.
Formula for electric field produced by a charge q.
E=
kq
r2
(this is a vector that points toward q if q < 0, or away otherwise)
E = electric field (N/C = Volts / meter)
q = charge (Coulomb)
r = distance away from q (meter)
k = 8.99 x 109 Newton-meters2/Coulomb2
Example.
What is the electric field 3 meters away from a 2 Coulomb charge?
(8.99 ×109 Nm 2 / C 2 ) × (2C )
E=
(3m)2
= 2.0 x 109 N/C = 2.0 x 109 Volts/meter
What is the force that this electric field exerts on an electron?
(Electric Field) = Force / Charge
So
Force = (Electric Field) x Charge
= (2.0 x 109 N/C) x (1.6 x 10-19 C)
= 3.2 x 10-10 N