From mechanics we know that, between two bodies, which are not in contact with each other, there is a small gravitational **attraction force**.

Between two objects with **electrical charge**, depending on the sign of the charges, positive or negative, the interaction force can be of **attraction** or **repulsion**.

Let’s consider two electrical charges *q _{1}* and

*q*, separated by the distance

_{2}*r*. Between the electrical charges there is an interaction force which is attractive if the charges have opposite signs and repulsive if both charges have the same sign (either positive or negative).

The **Coulomb force **(*F*)*, *also called **electrostatic force** or **Coulomb interaction**, states that the magnitude of the electrostatic force of interaction between two point electrical charges (*q _{1}*,

*q*) is directly proportional to the scalar multiplication of the magnitudes of electrical charge and inversely proportional to the square of the distance (

_{2}*r*) between them.

The Coulomb force is along the straight line joining them. If the two electrical charges have the same sign, the electrostatic force between them is **repulsive**; if they have different signs, the force between them is **attractive**.

The mathematical expression of Coulomb’s law is:

\[\bbox[#FFFF9D]{F = k \frac{q_1 q_2}{r^2}} \tag{1}\]where:

*F [N]* – Coulomb force

*q _{1}, q_{2} [C]* – electrical charges

*r [m]*– distance between electrical charges

*k [F/m]*– is called the

**Coulomb’s constant**, or electric force constant or electrostatic constant.

The value of Coulomb’s constant is calculated as:

\[k = \frac{1}{4\pi\varepsilon_0} \tag{2}\]where *ε _{0}* is the

**electrical permittivity**of free space (vacuum).

The electrical permittivity is a constant, of value:

\[\varepsilon_0 = 8.854187817 \cdot 10^{-12} \quad \frac{F}{m}\]The unit of measurement for electrical permittivity is *Farad* per *meter*.

Replacing the value of electrical permittivity in equation (2), we can calculate the value of Coulomb’s constant:

\[k = 8.987552 \cdot 10^9 \quad \frac{Nm^2}{C^2}\]If the electrical charges are placed in another medium, water for example, instead of using the permittivity of vacuum, we need to use the **absolute permittivity** *ε* which is the product between the **permittivity of vacuum** *ε _{0}* and

**relative permittivity**ε

_{r}.

In this case, the Coulomb constant will be:

\[k = \frac{1}{4\pi\varepsilon} \tag{4}\]**Example**. Calculate the Coulomb force between two electrons in water, at the distance of 1 mm from each other.

q_1 &= q_2 = 1.6 \cdot 10^{-19} \quad C\\

r &= 0.001 \quad m\\

\varepsilon_0 &= 8.854187817 \cdot 10^{-12}

\end{split} \end{equation*} \]

The **relative permittivity of water** at 20 °C is:

The Coulomb force will be:

\[ \begin{equation*} \begin{split}F &= \frac{1}{4\pi\varepsilon} \cdot \frac{q_1 q_2}{r^2}\\

&=\frac{1}{4 \cdot \pi \cdot 8.854187817 \cdot 10^{-12} \cdot 80.1} \cdot \frac{1.6 \cdot 10^{-12} \cdot 1.6 \cdot 10^{-12}}{0.001^2}\\

&= 0.00000000028724260\\

&= 2.872426 \cdot 10^{-10} \quad N

\end{split} \end{equation*} \]

In order for Coulomb’s law to be valid, several conditions need to be fulfilled:

- the charges must have a spherically symmetric distribution
- the charges must not be in contact
- the charges must be stationary with respect to each other

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## Musty

Thanks for the write-up

My question is, what if the electrical charges are placed inside/on the ground? what would be the ground relative permittivity?