Coulomb's Law: Electric Force Between Charges

What force does a charged particle with a charge of \( 0.001 \, \text{C} \) experience when it is \( 0.2\, \text{m} \) away from a sphere with a charge of \( 0.5 \, \text{C} \)?

Solution to Exercise #1

We need to find out the electric force \(F_{\text e}\) between a charged particle with charge \( q_1 = 0.001 \, \text{C} \) and a sphere with charge \( q_2 = 0.5 \, \text{C} \) at a distance \( r = 0.2\, \text{m} \) from each other.

Here you don't need to rearrange Coulomb's Law, but you can directly substitute the given values. Here, the Coulomb constant \( \frac{1}{4\pi \varepsilon_0} \) corresponds to the value \( 8.987 \cdot 10^9 \, \frac{ \text{Vm} }{ \text{As} } \): 1-1 \begin{align} F_{\text e} &~=~ 8.987 \cdot 10^9 \, \frac{ \text{Vm} }{ \text{As} } ~\cdot~ \frac{0.001 \, \text{C} ~\cdot~ 0.5 \, \text{C}}{(0.2\, \text{m})^2} \\\\ &~=~ 1.1 \cdot 10^8 \, \text{N} \end{align}

So, the particle experiences a very large force of \(1.1 \cdot 10^8 \, \text{N}\).

What is the attraction force between the nucleus of an iron atom with a charge of \( 26 \, e \) and an electron at a distance of \( 10^{-12} \, \text{m} \)?

Solution to Exercise #2

The unknown force is the attractive electric force \( F_{\text e} \) between the nucleus of an iron atom, which carries the charge \( q_1 = 26\,e\), and an electron with charge \(q_2 = e \) at a distance \( r = 10^{-12} \, \text{m}\).

The elementary charge has the value \( e = 1.602 \cdot 10^{-19} \, \text{C}\). Thus, the iron nucleus has the charge: 2 \begin{align} q_1 &~=~ 26 ~\cdot~1.602 \cdot 10^{-19} \, \text{C} \\\\ &~=~ 4.168 \cdot 10^{-18} \, \text{C} \end{align}

With the Coulomb constant, the electric force is: 2-1 \begin{align} F_{\text e} &~=~ 8.987 \cdot 10^9 \, \frac{ \text{Vm} }{ \text{As} } ~\cdot~ \frac{4.168 \cdot 10^{-18} \, \text{C} ~\cdot~ 1.602 \cdot 10^{-19} \, \text{C} }{ \left(10^{-12} \, \text{m}\right)^2 } \\\\ &~=~ 0.006 \, \text{N} \end{align}

So, the nucleus of an iron atom exerts a force of \(0.006 \, \text{N}\) on the observed electron.

With what force do two protons in the atomic nucleus repel each other, assuming they are \( 5\cdot 10^{-15} \, \text{m} \) apart?

Solution to Exercise #3

The unknown force is the electric repulsion force \(F_{\text e}\) between two protons at a distance \( r = 5\cdot 10^{-15} \, \text{m} \). A proton carries the elementary charge \( q_1 = q_2 = 1.602 \cdot 10^{-19} \, \text{C}\).

With the Coulomb constant (see part (a)), the electric force is: 3 \begin{align} F_{\text e} &~=~ 8.987 \cdot 10^9 \, \frac{ \text{Vm} }{ \text{As} } ~\cdot~ \frac{ \left( 1.602 \cdot 10^{-19} \, \text{C} \right)^2 }{ \left( 5\cdot 10^{-15} \, \text{m} \right)^2 } \\\\ &~=~ 9.22 \, \text{N} \end{align}

So, two protons in an atomic nucleus repel each other with a force of \(9.22 \, \text{N}\).

Two equally charged spheres repel each other with a force of \( 0.2 \, \text{N} \) when their distance from each other is \( 1\, \text{m} \). What charge do these spheres carry?

Solution to Exercise #4

The task is to find the charge \(q\) of the spheres. Both carry the same charge. That means \(q = q_1 = q_2\). Given is their repulsion force \( F_{\text e} = 0.2 \, \text{N} \) at a distance \( r = 1 \, \text{m} \) from each other. Thus, we need to rearrange Coulomb's Law for the charge \(q\). Bring \(r^2\) to the left side: 4.1 \[ F_{\text e} \, r^2 ~=~ \frac{1}{4\pi \varepsilon_0} \, q^2 \]

Bring \(4\pi \varepsilon_0\) to the left side: 4.2 \[ 4\pi \varepsilon_0 \, F_{\text e} \, r^2 ~=~ q^2 \]

Take the square root on both sides: 4.3 \[ \sqrt{ 4\pi \varepsilon_0 \, F_{\text e} \, r^2 } ~=~ q \]

Substitute the given values: 4.4 \begin{align} q &~=~ \sqrt{ 4\pi \varepsilon_0 \, F_{\text e} \, r^2 } \\\\ q &~=~ \sqrt{ 4\pi ~\cdot~ 8.854 \cdot 10^{-12} \, \frac{ \text{As} }{ \text{Vm} } ~\cdot~ 0.2 \, \text{N} ~\cdot~ \left( 1 \, \text{m} \right)^2 } \\\\ q &~=~ 4.7 \cdot 10^{-6} \, \text{C} \end{align}

The charged spheres repel each other with a force of \(4.7 \,\mu\text{C}\).

Two fixed, charged spheres, both with a charge of \( 10^{-6} \, \text{C} \), exert a force of \( 1 \, \text{N} \) on each other. At what distance are they from each other?

Solution to Exercise #5

To do this, rearrange Coulomb's law for the distance \( r \). Bring \( r^2 \) to the left side: 5.1 \[ F_{\text e} \, r^2 ~=~ \frac{q^2}{4\pi \varepsilon_0} \]

Then bring \( F_{\text e} \) to the right side: 5.2 \[ r^2 ~=~ \frac{q^2}{4\pi \varepsilon_0 \, F_{\text e}} \]

Take the square root on both sides: 5.3 \[ r ~=~ \sqrt{ \frac{q^2}{4\pi \varepsilon_0 \, F_{\text e}} } \]

Now plug in the given values: 5.4 \begin{align} r &~=~ \sqrt{ \frac{ \left( 10^{-6} \, \text{C} \right)^2 }{ 4\pi ~\cdot~ 8.854 \cdot 10^{-12} \, \frac{ \text{As} }{ \text{Vm} } ~\cdot~ 1 \, \text{N} } } \\\\ &~=~ 8.99 \cdot 10^{-3} \, \text{m} \end{align}

The two charged spheres are approximately \( 9 \, \text{mm} \) apart from each other.

You want to experimentally determine the electric field constant using Coulomb's law. To do this, you charge two spheres equally with \( 2 \cdot 10^{-8} \, \text{C} \), place them \( 0.01\, \text{m} \) apart from each other, and measure a force of \( 0.036\, \text{N} \). What is the value of the electric field constant in this measurement?

Solution to Exercise #6

Given is the charge \( q_1 = q_2 = 2 \cdot 10^{-8} \, \text{C} \), the distance \( r = 0.01\, \text{m} \), and the force \( F_{\text e} = 0.036\, \text{N} \).

Rearrange Coulomb's law for \( \varepsilon_0 \). Multiply both sides by \( \varepsilon_0 \): 6.1 \[ \varepsilon_0\, F_{\text e} ~=~ \frac{1}{4\pi } \, \frac{q^2}{r^2} \]

Multiply both sides by \( \frac{1}{ F_{\text e} } \): 6.2 \[ \varepsilon_0 ~=~ \frac{1}{4\pi F_{\text e}} \, \frac{q^2}{r^2} \]

Plug in the given values: 6.3 \begin{align} \varepsilon_0 &~=~ \frac{1}{4\pi ~\cdot~ 0.036\, \text{N} } ~\cdot~ \frac{ \left(2 \cdot 10^{-8} \, \text{C}\right)^2 }{(0.01\, \text{m})^2} \\\\ &~=~ 8.84 \cdot 10^{-12} \, \frac{ \text{As} }{ \text{Vm} } \end{align}

Two sodium spheres, each with a mass \(M = 0.001 \, \text{kg} \), are positioned at a distance \( r = 1 \, \text{m} \) from each other. One-tenth of the sphere consists of simply positively charged \(\text{Na}^+\) ions (i.e., one valence electron has been removed from a sodium atom).

What is the magnitude of the electric force \(F_{\text e}\) between the two sodium spheres?

Hint: The mass of a sodium atom is: $$ m ~=~ 23 ~\cdot~ 1.67 \cdot 10^{-27} \, \text{kg} $$

Solution to Exercise #7

To solve the problem, use Coulomb's law. The goal is simply to determine the charges \(q_1\) and \(q_2\) of the two sodium spheres. The distance between the spheres is known: \( r = 1 \, \text{m} \).

Since the two sodium spheres are identical in size and especially since they both lack the same number of electrons, they carry the same charge: \( q:= q_1 = q_2 \). Simply denote them as \(q\). Coulomb's law 1 thus becomes: \[ F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{q^2}{r^2} \]

Now, only one unknown quantity, \(q\), needs to be determined. It is known that one-tenth of all sodium atoms in a sphere lack a valence electron. A valence electron carries the negative charge \( e = -1.602 \cdot 10^{-19} \, \text{C} \), so one-tenth of the sodium atoms are positively charged (they are called \(\text{Na}^+\) ions). Thus, a single \(\text{Na}^+\) ion carries the charge \( e = +1.602 \cdot 10^{-19} \, \text{C} \). The total charge \(q\) of the sphere is simply the number of \(\text{Na}^+\) ions \(N_{\text e}\) multiplied by the charge \(e\): \( q = N_{\text e} \, e\): \[ F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{\left(N_{\text e} \, e\right)^2}{r^2} \]

However, the only thing known about their number is that it is one-tenth of the total number \(N\) of all sodium atoms in the sphere: \(N_{\text e} = \frac{1}{10} \, N \): 5 \[ F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{\left( \frac{1}{10} \, N \,e \right)^2}{r^2} \]

To determine how many sodium atoms make up a sphere, the mass of the sphere must be known (it is: \(M = 0.001 \, \text{kg} \)) and the mass of a single sodium atom. The latter can be read from the periodic table of elements: \(m = 23 \cdot 1.67 \cdot 10^{-27} \text{kg} \). The ratio of the masses corresponds to the number of sodium atoms in the sphere: 6 \[ N ~=~ \frac{M}{m} \]

Simply substitute Eq. 6 into Eq. 5: 7 \[ F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{\left( \frac{1}{10} \, \frac{M}{m} \,e \right)^2}{r^2} \]

Nice, all quantities in 7 are known. Substitute them to find out with what force the two small spheres repel each other: 8 \[ F_{\text e} ~=~ 1.56 \cdot 10^{15} \, \text{N} \]

The calculated force is not a computational error, but actually so unimaginably large. For comparison: The Earth and the Moon attract each other with a force of \( 10^{20} \, \text{N}\).

Determine the potential energy \(W_{\text{pot}}\) contained in the following charge arrangements:

1. Triangular charge arrangement with point charges \(Q_1\), \(Q_2\), and \(Q_3\). All charges are at a distance \(d\) from each other.
2. Quadrupole arrangement with two point charges \(Q\) and two point charges \(-Q\) spaced at distance \(d\).

Solution to Exercise #8.1

To calculate the potential energy, the superposition principle is applied. This principle states that the total electric force \(F\) on a test charge, exerted by any charge distribution, is given by the sum of the forces \(F_1, F_2,~...\) of all individual source charges exerting a force on the test charge: \[ F ~=~ F_1 + F_2 + ... + F_n \]

This property also applies to electric fields, potentials and potential energies. Therefore, in the following, the triangular arrangement is built up charge by charge to form the sum of potential energies.

Since point charges are considered here, the electric potential emanating from the charge \(Q_1\) is given by: \[ V(r) ~=~ \frac{Q_1}{4\pi \, \varepsilon_0} \, \frac{1}{r} \]

The potential \(V\) denotes the potential energy per charge at location \(r\).

When another charge \(Q_2\) is placed at distance \(r\) from \(Q_1\), then this charge \(Q_2\) has the following potential energy: \[ W_1 ~=~ Q_2 \, V(r) ~=~ Q_2 \, \frac{Q_1}{4\pi \, \varepsilon_0} \, \frac{1}{r} \]

According to the task, the charge \(Q_2\) is placed at a distance \(d\) from the charge \(Q_1\). So: 2 \[ W_1 ~=~ Q_2 \, \frac{Q_1}{4\pi \, \varepsilon_0} \, \frac{1}{d} \]

Now, it's a new charge arrangement: The total potential is now given not only by a charge \(Q_1\) but also by the other charge \(Q_2\). Therefore, to determine the potential energy of the third charge \(Q_3\), first, the potential dependent on both \(Q_1\) and \(Q_2\) must be calculated. And that's not difficult because the principle of superposition allows first to calculate the potential energy that another charge \(Q_3\) will have when placed at a distance \(d\) from \(Q_1\): 3 \[ W_2 ~=~ Q_3 \, \frac{Q_1}{4\pi \, \varepsilon_0} \, \frac{1}{d} \]

And then the potential energy of \(Q_3\) when placed at a distance \(d\) from \(Q_2\): 4 \[ W_3 ~=~ Q_3 \, \frac{Q_2}{4\pi \, \varepsilon_0} \, \frac{1}{d} \]

The total potential energy \(W_{\text{pot}}\) of this triangular arrangement is thus the sum of 2, 3, and 4: 5 \[ W_{\text{pot}} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{1}{d} \, \left( Q_1 \, Q_2 + Q_1 \, Q_3 + Q_3 \, Q_2 \right) \]

For example, if \(Q_1 = Q_2 = Q_3 = e\), then 5 becomes: 6 \[ W_{\text{pot}} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{3 e^2}{d} \]

So, this energy is present in the system of three positive elementary charges \(e\), each at a distance \(d\) from each other. If these charges are not held together at this distance \(d\) by an external force, the triangular arrangement is unstable. The charge carriers will convert their potential energy into kinetic energy through mutual repulsion.

On the other hand, if \(Q_1 = -e\) and \(Q_2 = Q_3 = e\), then 5 becomes: 7 \[ W_{\text{pot}} ~=~ -\frac{1}{4\pi \, \varepsilon_0} \, \frac{e^2}{d} \]

This system has a negative total potential energy. This system is stable, and this energy would be necessary to destroy the system.

Solution to Exercise #8.2

The procedure follows the same steps as in Exercise #8.1. The charge \(-Q\) causes a negative potential at distance \(d\), given by: 8 \[ V(d) ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{-Q}{d} \]

When a positive charge \(+Q\) is brought to a distance \(d\) from \(-Q\), the system has the potential energy: 9 \[ W_1 ~=~ Q \, \frac{1}{4\pi \, \varepsilon_0} \, \frac{-Q}{d} \]

This is the energy of an electric dipole consisting of a positive and negative charge. A third charge \(-Q\), brought as in the diagram to form the dipole, then has the potential energy: 10 \[ W_2 ~=~ -Q \, \frac{1}{4\pi \, \varepsilon_0} \, \left( \frac{Q}{d} + \frac{-Q}{\sqrt{2} \, d} \right) \]

In Eq. 10, it must be noted that \(-Q\) at a distance from the other \(-Q\) charge (see diagram) does not have the distance \(d\) but the distance given by the diagonal. This can be easily found using the Pythagorean theorem: \(\sqrt{d^2 + d^2}\).

To obtain a quadrupole, the last positive charge \(+Q\) is added to the charge arrangement of three charges. The new charge then has the following potential energy: 11 \[ W_3 ~=~ Q \, \frac{1}{4\pi \, \varepsilon_0} \, \left( \frac{-Q}{d} + \frac{-Q}{d} + \frac{Q}{\sqrt{2} \, d} \right) \] Here too, the diagonal distance between \(+Q\) and \(+Q\) must be considered.

The sum of 9, 10, and 11 gives the total potential energy of the quadrupole: \[ W_{\text{pot}} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q^2}{d} \left( -4 + \sqrt{2} \right) \]

Here, \( \frac{2}{\sqrt{2}} = \sqrt{2} \) was used. Expand the bracket and you get: \[ W_{\text{pot}} ~\approx~ -\frac{2.6 \, Q^2}{4\pi \, \varepsilon_0 \, d} \]