Coulomb's Law: Electric Force Between Charges
Basics for understanding Coulomb's law
There are two different types of electric charge:

positive charge

negative charge
Why specifically two types of charges and not one or even three? Because the existing experiments (such as the deflection of charges in electric and magnetic fields) could identify precisely TWO types of charges that differ in their force interaction with each other. Positive charges repel each other with a certain force. Negative charges also repel each other. However, a positive and a negative charge attract each other.
Electric charge is abbreviated with a small \( q \). And the physical unit of electric charge has been defined as the unit »Coulomb« (\(\text C \)), which can also be written as an amperesecond (\(\text{As} \)):
Remember that knowing the units can help you to derive other units or to check the correctness of your transformations of formulas.
With what force do charges attract each other?
You now know that there are two different types of charge that exert a repulsive or attractive electric force \( F_{\text e} \) on each other. You probably have a few questions now: How can I calculate this force? What if the charges are of different sizes? What if the distance between them changes? All these questions can be answered in an experiment. Coulomb's law is a typical physical law that was originally discovered by experiment and not by mathematical derivation.
In general, the experiment for Coulomb's Law proceeds as follows: You electrically charge two metallic spheres so that they carry an electric charge, which you prescribe, for example, using a voltage source. Then, you measure the force between the spheres at a specific distance \( r \) apart. You vary both the distance (that is, \(0.1\, \mathrm{m}, 0.2\, \mathrm{m}, 0.3\, \mathrm{m}\)...) and the charges on one sphere \( q_1 \) and the second sphere \( q_2 \) (that is, \(0.5\, \mathrm{C}, 0.6\, \mathrm{C}, 0.7\, \mathrm{C}\)...). A specific experiment for determining Coulomb's Law is, for instance, the Coulomb's Torsion Balance.
All the measured values you have collected are then examined further in a diagram. In a \(F_{\text e}\)\(r\)diagram (that is, force as a function of the charge distance) you find out that the electric force between two charges \(q_1 \) and \(q_2 \), which carry the two spheres, is proportional to \( \frac{1}{r^2} \):
This means: If you double the distance \(r\), the force \( F_{\text e} \) is reduced by a factor of FOUR!
Then look at how the force changes when you use different charge values. To do this, vary the charge of a sphere. Plot the measured values in a \(F_{\text e}\)\(q_1\) diagram. The result is a linear relationship:
If you vary the charge \( q_2 \) of the second sphere, you will of course get the same proportionality:
If you combine the three experimental relationships 1
, 2
, and 3
, you will already get the following proportionality:
Only the proportionality constant \( K \) needs to be determined in order to fully establish Coulomb's law:
By measuring the force between two (known) charges and their distance from each other, you can find the required constant \( K \) by rearranging the equation 5
. For example, charge the two spheres so that they have \( q_1 = q_2 = 10^{4} \, \mathrm{C} \) and place them at a distance \( r = 1 \, \mathrm{m} \) from each other. Then you will measure a force \( F_{\text e} = 89.875 \, \mathrm{N} \) between the two charges. Rearrange 5
with respect to \( K \) and enter the measured value provided here:
~&=~ \frac{89.875 \, \mathrm{N} \cdot 1 \, \mathrm{m}^2}{10^{4} \, \mathrm{C} \cdot 10^{4} \, \mathrm{C}} \\\\
~&=~8.9875 \cdot 10^9 \, \frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{C}^2}
Of course, it doesn't matter which charges and which distance you choose, the result for the proportionality constant will always be the same because it is a constant!
Later on in your physics adventures it will turn out that it makes sense to define the Coulomb constant as follows:
The value of the vacuum permittivity \( \varepsilon_0 \) can be determined by rearranging Eq. 7
for \( \varepsilon_0 \) and then substituting the value of \(K\) (Eq. 6
):
~&=~ \frac{1}{ 4 \cdot 3.1415 \cdot 8.9875 \cdot 10^9 \, \frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{C}^2}} \\\\
~&=~8.854 \cdot 10^{12} \, \frac{\mathrm{C}^2}{\mathrm{N}\,\mathrm{m}^2}
The unit of the vacuum permittivity is usually given in amperesecond by voltmeter. This can easily be done by converting the units:
&~=~ \frac{\mathrm{A}^2 \,\mathrm{s}^4}{\mathrm{kg} \, \mathrm{m}^3} \\\\
&~=~ \frac{\mathrm{As}}{\mathrm{Vm}}
The vacuum permittivity \( \varepsilon_0 \) is encountered nearly everywhere electricity and magnetism are present because this fundamental constant dictates the strength of the electromagnetic interaction between electric charges, shaping the universe as we know it. Why precisely this value 8
is prescribed by nature and not another can only be answered by a "higher power."
The journey ends here, because if you summarize everything you have learned so far, you will come to the following physical relationship, which is called Coulomb's law in honour of the French physicist Charles Augustin de Coulomb, who experimented a lot with charges:
Is the Electric Force Repulsive or Attractive?
Depending on whether \( q_1 \) or \( q_2 \) is positive or negative in Coulomb's Law 10
, the electric force \( F_{\text e} \) has a different sign, resulting in a repulsive or attractive effect between the two charges:

\( q_1\) and \(q_2 \) are both positive. Then \( F_{\text e} \) is also positive.

\(q_1\) and \(q_2 \) are both negative. Then \( F_{\text e} \) is positive, because "" times "" equals "+".

\(q_1\) is positive and \(q_2\) is negative (or vice versa). Then \( F_{\text e} \) is negative, because "+" times "" equals "".
Example: Force between electron and proton
According to the simplest atomic model of the hydrogen atom, a negatively charged electron orbits the atomic nucleus, which consists of a single positively charged proton. Both the electron and the proton carry the elementary charge:
According to the model, the radius \(r\) of the electron orbit is:
What is the electric force \( F_{\text e} \) that the electron and proton exert on each other at this distance? Do they attract or repel each other?
Use Coulomb's law 10
to find the electric force between the two charges:
~&=~ \frac{1}{4\pi \cdot 8.85 \cdot 10^{12} \, \frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{C}^2}} \cdot \frac{1.60 \cdot 10^{19} \, \mathrm{C} \cdot 1.60 \cdot 10^{19} \, \mathrm{C}}{(0.53 \cdot 10^{10} \, \mathrm{m})^2} \\\\
~&=~ 8.19 \cdot 10^{8} \, \mathrm{N}
Since the force is negative, the electron and proton attract each other!
Coulomb's law in a medium
Coulomb's law 10
only applies in a vacuum or approximately in air. If you place the two charged spheres in water (not salty water, of course, otherwise there will be a short circuit), you will find that the medium between the charges also plays a crucial role and we have not taken it into account! To take other media into account, the Coulomb law becomes:
Here, \( \varepsilon_{\text r} \) is the introduced relative permittivity (also called dielectric constant). This dimensionless quantity is used to account for the medium between the charges.
 If the two charges are in vacuum: \( \varepsilon_{\text r} = 1 \).
 If the two charges are in the air: \( \varepsilon_{\text r} = 1.0006 \approx 1 \).
 If the two charges are in lukewarm water: \( \varepsilon_{\text r} \approx 81.1 \).
Electric Potential and Potential Energy
An electric charge \(\class{red}{Q}\) at a distance \(r\) generates an electric potential \(V(r)\), given by the following formula: \[ V(r) ~=~ \frac{\class{red}{Q}}{4\pi \, \varepsilon_0} \, \frac{1}{r} \]
The electric potential \(V\) indicates the potential energy per charge. The potential energy \( W_{\text{pot}}(r) \) of a charge \(\class{red}{q}\), which is placed at a distance \(r\) from the source charge \(\class{red}{Q}\), is then given by $$\begin{align} W_{\text{pot}}(r) ~&=~ \class{red}{q} \, V(r) \\\\ ~&=~ \frac{\class{red}{q} \, \class{red}{Q}}{4\pi \, \varepsilon_0} \, \frac{1}{r} \end{align}$$
Exercises with Solutions
Use this formula eBook if you have problems with physics problems.With the following exercises, you practice manipulating the Coulomb's Law.
Exercise #1: A Charged Particle
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 \( 8.987 \cdot 10^9 \, \frac{ \text{Vm} }{ \text{As} } \): 11 \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 \(1.1 \cdot 10^8 \, \text{N}\).
Exercise #2: The Iron Atom
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: 21 \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.
Exercise #3: The Atomic Nucleus
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 \(9.22 \, \text{N}\).
Exercise #4: Charge of the Spheres
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 \(4.7 \,\mu\text{C}\).
Exercise #5: Distance between the Spheres
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.
Exercise #6: Determining the Electric Field Constant
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}
Exercise #7: Electric Force between Two Sodium Spheres
Two sodium spheres, each with a mass \(M = 0.001 \, \text{kg} \), are positioned at a distance \( r = 1 \, \text{m} \) from each other. Onetenth 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 onetenth 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 onetenth 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 onetenth 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 \( 10^{20} \, \text{N}\).
Exercise #8: Charges in a Triangular and Quadrupole Arrangement
Determine the potential energy \(W_{\text{pot}}\) contained in the following charge arrangements:
 Triangular charge arrangement with point charges \(Q_1\), \(Q_2\), and \(Q_3\). All charges are at a distance \(d\) from each other.
 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: \[ W_{\text{pot}} ~\approx~ \frac{2.6 \, Q^2}{4\pi \, \varepsilon_0 \, d} \]