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8.1  Coulomb's Law
 To know that there exists a force between two charges and that force can be either attractive or repulsive.
 To know and be able to use the inversesquare law of electrostatics (Coloumb's Law)
 To appreciate the similarities and differences between Coulomb's Law and Gravitation.
 To be able to solve problems involving two charged particles.
Force between two electrically charged particles is given by Coulomb's Law
\[F_{e} = k\frac{q_{1} q_{2}}{r^2}\]
where \(k\) is the Coulomb constant that is a function of the permittivity of the space between the charges. The permittivity of air is almost identical to the permittivity of free space (vacuum), \(\epsilon_o\).
\[\epsilon_{o} = 8.85\times 10^{12} \,\text{C}^2 \text{/Nm}^2\]
\[k=\frac{1}{4 \pi \epsilon_{o}} = 8.99 \times 10^{9} \, \text{Nm}^2 \text{/C}^2 \]
Personally, I grew up using the original form and I prefer it, but the AP examiners are happy for you to use \(k\), which is quicker.
\[F_{e} = k\frac{q_{1} q_{2}}{r^2}\]
where \(k\) is the Coulomb constant that is a function of the permittivity of the space between the charges. The permittivity of air is almost identical to the permittivity of free space (vacuum), \(\epsilon_o\).
\[\epsilon_{o} = 8.85\times 10^{12} \,\text{C}^2 \text{/Nm}^2\]
\[k=\frac{1}{4 \pi \epsilon_{o}} = 8.99 \times 10^{9} \, \text{Nm}^2 \text{/C}^2 \]
Personally, I grew up using the original form and I prefer it, but the AP examiners are happy for you to use \(k\), which is quicker.
Experimental Work
Coulomb's Law leads to some pretty weak forces between the charges that we are able to produce in the lab. The classic experiment is to suspend two very light pith balls that have been painted with conducting material from lengths of cotton. Once they are charged up they repel and the force between them can be deduced from the angle that they stabilize at. This requires the use of statics from Unit 2. It is next to impossible to measure the charges on the spheres, so instead we calculate the product of the charges and see if they are realistic. Usually microcoulombs. To do a proper investigation that can yield a graph and a value for the electrostatic constant, \(k\), or the permittivity of free space, \(\epsilon_{o}\), we have to resort to a computer simulation. An excellent one has been created by the everresourceful Physics Aviary website.
Coulomb's Law leads to some pretty weak forces between the charges that we are able to produce in the lab. The classic experiment is to suspend two very light pith balls that have been painted with conducting material from lengths of cotton. Once they are charged up they repel and the force between them can be deduced from the angle that they stabilize at. This requires the use of statics from Unit 2. It is next to impossible to measure the charges on the spheres, so instead we calculate the product of the charges and see if they are realistic. Usually microcoulombs. To do a proper investigation that can yield a graph and a value for the electrostatic constant, \(k\), or the permittivity of free space, \(\epsilon_{o}\), we have to resort to a computer simulation. An excellent one has been created by the everresourceful Physics Aviary website.
WORKED EXAMPLE
In 1910, Rutherford organised an experiment where he bombarded a thin sheet of gold foil with high energy alpha particles. The surprising discovery that some of the alpha particles bounced back led to the discovery of the atomic nucleus. (See IGCSE Physics  Nuclear Physics for more details). To get a value for the force of repulsion that the alpha particle received from the gold nucleus we can do the following calculation. An alpha particle (2 protons and 2 neutrons) is sent at high speed toward a gold nucleus (79 protons). What is the electrical force acting of the alpha particle when it is \(2.0 \times 10^{14}\,\text{m}\) from the gold nucleus? (2) \[F=k \frac{q_1 q_2}{r^2}\] \[F=\frac{8.99 \times 10^9 \times (2 \times 1.6 \times 10^{19} ) \times (79 \times 1.6 \times 10^{19})}{(2.0 \times 10^{14})^2} \] \[F = 90.9 \,\text{N}\] which is a HUGE force to apply to a subatomic particle. The acceleration of the alpha particle can be calculated: \[a=\frac{F}{m} = \frac{90.9}{(4 \times 1.6 \times 10^{27})} \] \[a = 1.4 \times 10^{28} \,\text{m/s}^2\] 
WORKED EXAMPLE
The classical model of the hydrogen atom has an electron in orbit around a proton. An obvious question is  how fast does it go? We can use Coulomb's Law to give us an order of magnitude. The mass of the electron is \(9.11 \times 10^{31}\,\text{kg}\) and the radius of the hydrogen atom is \(r\approx 0.5 \times 10^{10}\,\text{m}\). 
Concept: the electrostatic attraction provides the centreseeking force required for circular motion.
\[F_e = F_c\]
\[k\frac{e^2}{r^2} = \frac{mv^2}{r}\]
a bit of algebra gives us:
\[v^2 = \frac{ke^2}{mr}\]
\[v = \sqrt{\frac{ (8.99 \times 10^9 \times (1.6\times 10^{19})^{2}) }{ (9.11 \times 10^{31} \times 0.5 \times 10^{10} )}}\]
\[v = 2.24 \times 10^{6} \,\text{m/s}\]
(hint: a good check is that if you ever have an answer that is faster than the speed of light  it is wrong! Normally as you forgot to square root it)
\[F_e = F_c\]
\[k\frac{e^2}{r^2} = \frac{mv^2}{r}\]
a bit of algebra gives us:
\[v^2 = \frac{ke^2}{mr}\]
\[v = \sqrt{\frac{ (8.99 \times 10^9 \times (1.6\times 10^{19})^{2}) }{ (9.11 \times 10^{31} \times 0.5 \times 10^{10} )}}\]
\[v = 2.24 \times 10^{6} \,\text{m/s}\]
(hint: a good check is that if you ever have an answer that is faster than the speed of light  it is wrong! Normally as you forgot to square root it)
WORKED PAST AP B QUESTION (modified)
Surprisingly, there has not been a Coulomb's Law FRQ since the AP1 exam started in 2015. Hmmm.... this could be the year!
Surprisingly, there has not been a Coulomb's Law FRQ since the AP1 exam started in 2015. Hmmm.... this could be the year!
A charge \(Q_{1} = –16.0 \times 10^{–6} \,\text{C}\) is fixed on the \(x\) axis at \(+4.0\,\text{m}\), and a charge \(Q_2 = +9.0\times 10^{–6}\,\text{C}\) is fixed on the \(y\) axis at \(+3.0\,\text{m}\), as shown on the diagram.
A third charge of magnitude \(Q_3 = –4 \times 10^{–6}\,\text{C}\) is placed at the origin. Unlike the first two charges, it is free to move. 
a) Calculate the resultant force on the new charge due to the other, fixed charges.
Solution: There are two forces, one due to each pair of charges. We need to work these out, then add the vectors together to get the resultant force.
Solution: There are two forces, one due to each pair of charges. We need to work these out, then add the vectors together to get the resultant force.
Force between \(Q_1\) and \(Q_3\), which is the \(x\) direction
\[F_x = \frac{(8.99 \times 10^9 \times 4 \times 10^{–6} \times 16 \times 10^{–6})} {4^2}\] \[F_x = 0.036\,\text{N}\] 
Force between \(Q_2\) and \(Q_3\), which is the \(y\) direction
\[F_y = \frac{(8.99 \times 10^9 \times 4 \times 10^{–6} \times 9.0\times 10^{–6})} {3^2} \] \[F_y = 0.036\,\text{N}\] 
Combining these two forces (which nicely are of equal magnitude) using Pythagoras' Theorem gives us the resultant.
\[F = \sqrt{(0.036)^2 + (0.036)^2} = 0.051 \,\text{N}\]
Simple inspection of the which way the forces go (repelled by \(Q_1\) and attracted to \(Q_2\)) coupled with the symmetrical geometry tells you that the resultant force vector on \(Q_3\) is \(45^{\circ}\) to the top left. If the forces were not equal, you would use \(\tan{\theta} = \frac{O}{A}\) to calculate the angle.
(The rest of the question dealt with electric field and potentials, which is AP2 stuff)
Sample MC Questions
\[F = \sqrt{(0.036)^2 + (0.036)^2} = 0.051 \,\text{N}\]
Simple inspection of the which way the forces go (repelled by \(Q_1\) and attracted to \(Q_2\)) coupled with the symmetrical geometry tells you that the resultant force vector on \(Q_3\) is \(45^{\circ}\) to the top left. If the forces were not equal, you would use \(\tan{\theta} = \frac{O}{A}\) to calculate the angle.
(The rest of the question dealt with electric field and potentials, which is AP2 stuff)
Sample MC Questions
1. Two identical conducting spheres are charged to \(+2Q\) and \(–Q\). respectively, and are separated by a distance d (much greater than the radii of the spheres) as shown above. The magnitude of the force of attraction on the left sphere is \(F_1\). After the two spheres are made to touch and then are reseparated by distance d, the magnitude of the force on the left sphere is \(F_2\). Which of the following relationships is correct?
(A) \(2F_1 = F_2\) (B) \(F_1 = F_2\) (C) \(F_1 = 2F_2\) (D) \(F_1 = 8 F_2\)
2. Two isolated charges, \(+ q\) and \(– 2q\), are \(2\,\text{cm}\) apart. If \(F\) is the magnitude of the force acting on charge \(–2Q\), what are the magnitude and direction of the force acting on charge \(+q\)?
Magnitude Direction
(A) \(2 F\) Away from charge \(–2q\)
(B) \(F\) Toward charge \(– 2q\)
(C) \(F\) Away from charge \(– 2q\)
(D) \(2F\) Toward charge \(– 2q\)
3. Multiple correct: Forces between two objects which are inversely proportional to the square of the distance between the objects include which of the following? Select two answers:
A.. Gravitational force between two celestial bodies
B. Electrostatic force between two electrons
C. Nuclear force between two neutrons
D. Magnetic force between two magnets
4. Two small spheres have equal charges \(q\) and are separated by a distance \(d\). The force exerted on each sphere by the other has magnitude \(F\). If the charge on each sphere is doubled and \(d\) is halved, the force on each sphere has magnitude
(A) \(F\) (B) \(2F\) (C) \(8F\) (D) \(16F\)
(A) \(2F_1 = F_2\) (B) \(F_1 = F_2\) (C) \(F_1 = 2F_2\) (D) \(F_1 = 8 F_2\)
2. Two isolated charges, \(+ q\) and \(– 2q\), are \(2\,\text{cm}\) apart. If \(F\) is the magnitude of the force acting on charge \(–2Q\), what are the magnitude and direction of the force acting on charge \(+q\)?
Magnitude Direction
(A) \(2 F\) Away from charge \(–2q\)
(B) \(F\) Toward charge \(– 2q\)
(C) \(F\) Away from charge \(– 2q\)
(D) \(2F\) Toward charge \(– 2q\)
3. Multiple correct: Forces between two objects which are inversely proportional to the square of the distance between the objects include which of the following? Select two answers:
A.. Gravitational force between two celestial bodies
B. Electrostatic force between two electrons
C. Nuclear force between two neutrons
D. Magnetic force between two magnets
4. Two small spheres have equal charges \(q\) and are separated by a distance \(d\). The force exerted on each sphere by the other has magnitude \(F\). If the charge on each sphere is doubled and \(d\) is halved, the force on each sphere has magnitude
(A) \(F\) (B) \(2F\) (C) \(8F\) (D) \(16F\)
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