In mechanics, we already introduced the concept of potential energy. For a radial gravitational field, we derived the potential energy relative to a faraway point by integrating the gravitational force and setting the upper limit of the integral to infinity. The same principle can be applied in electrostatics if the force $\vec{F}$ can be expressed as an analytical function:
$$
E_\mathrm{pot} = \int_{R}^\infty \vec{F}\cdot d\vec{s}
$$
Substituting Coulomb’s law for the force gives the potential energy of two point charges relative to infinity:
$$
E_{\mathrm{pot}} = \int_{R}^\infty \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\,dr
= \frac{q_1 q_2}{4\pi R}
$$
To make the result independent of $q_2$, the electric potential is defined as the ratio of potential energy to charge:
$$
\boxed{\varphi = \frac{E_{\mathrm{pot}}}{q}}
$$
For a point charge, this becomes:
$$
\varphi = \frac{Q}{4\pi R}
$$
where $q_1$ has been replaced with $Q$ for clarity. This same result can also be obtained directly from the definition of the electric field:
$$
\varphi = \frac{E_\mathrm{pot}}{q} = \int_{R}^\infty \vec{E}\cdot d\vec{s}
$$
When moving a charge from point $A$ to point $B$, the potential energy relative to infinity is less useful than the potential difference between the two points. This potential difference is called the electric voltage:
$$
U = \Delta\varphi = \varphi_B - \varphi_A
= \int_{B}^{\infty} \vec{E}\cdot d\vec{s} - \int_{A}^{\infty} \vec{E}\cdot d\vec{s}
$$
The voltage, denoted by $U$, is measured in volts (V), named after Alessandro Volta. After simplifying, the definition of voltage becomes:
$$
\boxed{U = \int_A^B \vec{E}\cdot d\vec{s}}
$$
The potential energy or the work performed can then be written as:
$$
\boxed{E_\mathrm{pot} = qU}
$$
To determine the final velocity of a charge accelerated through a potential difference, we can use energy conservation:
$$
qU = \frac{1}{2}mv^2
$$
Solving for $v$ gives:
$$
\boxed{v = \sqrt{\frac{2qU}{m}}}
$$
This velocity depends only on the potential difference, not on the length of the acceleration path.
If an electron is accelerated through a potential difference of 1 V, it acquires a kinetic energy of about $1.6 \cdot 10^{-19}$ J. Since the joule is inconveniently large for such small values, the electronvolt (eV) is used instead. By definition, an electron gains 1 eV of energy when accelerated through a potential of 1 V. The conversion is:
$$
\boxed{E[\mathrm{eV}] = e \, E[\mathrm{J}]}
$$