In the first part about electric fields, the magnitude and the direction of the electric field and the electric force have been described. The second part is about electric work, energy and the movement of charges in an electric field.

From earlier grades the *mechanical work* is already known. Mechanical work is done when a force \(F \) acts on a body along a path \(s \). The following holds:

\( F \) = Force, \( s \) = Distance

When charges are moved in a electric field, either through external influences or through the electric force of the field itself, also work is being done. It should however be noted that the path \(s \) is the path * parallel to the field lines * in this case. When you move a charge along the lines work is done, when moving perpendicular to the field lines, however, no work is performed.

For ** homogeneous fields** you can use the following formula.

\begin{aligned}
W &= Q \cdot E \cdot s \\
& \\
W &= Q \cdot U
\end{aligned}
\( Q \) = Charge of a body, \( E \) = Electric field

\( U \) = Voltage between start point and end point, \( s \) = Path parallel to the field lines

The following animation shows 3 charges (\ (q_1 = e \), \(q_2 = -e \), \(q_3 = -e \)) in a homogeneous field and shows the path parallel to the field lines.

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Although charges 1 and 2 travel different paths, the path parallel to the field lines is the same and therefore the amount of work performed is also the same.

Charge 3 moves perpendicular to the field lines, as such the length of the path parallel to the field lines is 0 and no work is done.

In **inhomogeneous fields**, the calculation of the electric work is more difficult since the field strength is not constant during movement of charges.

In the radial field of a point charge, the field strength depends on the distance \(r \). One can determine the work by integrating over the electric field:

$$ W = \int \limits_{r_1}^{r_2} F(r) \,\,\mathrm{d}r $$ With the coulomb force, we get:
$$ W = \dfrac{Q_1 \cdot Q_2}{4 \pi \cdot \epsilon_0 \cdot \epsilon_r} \cdot \int \limits_{r_1}^{r_2} \dfrac{1}{r^2} \,\,\mathrm{d}r = \dfrac{Q_1 \cdot Q_2}{4 \pi \cdot \epsilon_0 \cdot \epsilon_r} \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) $$
\( Q_1 \) = Charge of the point charge, \( Q_2 \) = Charge of the body,

\( r_1 \) = Distance at start position, \( r_2 \) = Distance at end position

The following animation shows 2 charges (\(q_1 = q_2 = -e \)) in an inhomogeneous field and shows the path parallel to the field lines.

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It can be seen here that the work being done is only determined by the start and end point. Whether moving the charges straight or with many curves is not important.

The potential energy is already known. In the gravitational field of the earth it is greater, the more a body is removed from the center of the earth, that is the higher it is above the earth's surface.

The following animation shows the analogy between the potential energy in the gravitational field and the potential energy in the electric field.

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In the gravitational field in the figure on the left has been set that the potential energy on the Earth's surface is 0. If one raises a body to a certain height, the work done is stored in the potential energy of the body.

Quite similarly, it is in the homogeneous electric field in the figure on the right. There it was defined that the potential energy of the negatively charged plate is 0. As you move the charged body, the work performed is stored in the potential energy of the body.

In an inhomogeneous radial field around a point charge, it is usually defined that the potential energy is 0 at an infinite distance from the point charge.

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Every place in an electric field has a specific **electric potential** \(\ phi \). This is a measure of how high is the potential energy of a charged body at this location.

$$ E_{pot} = \phi \cdot q $$ \( E_{pot} \) = Potential energy, \( \phi \) = Electric potential, \( q \) = Charge of the body

Stellt man diese Formel nach \( \phi \) um, so kann man die Formel für das Potential aus der Formel für die potentielle Energie herleiten, indem man einfach das \( q \) wegkürzt. Solving the formula for \(\ phi \), we observe that the formula for the potential can be derived from the formula for the potential energy by canceling \ (q \).

$$ \phi = \dfrac{E_{pot}}{q} $$
For the *homogeneous* field:

For the *inhomogeneous* field:

Similar to the field lines we can draw so-called potential lines in an electric field. They run perpendicular to the field lines. During the movement of a body along a potential line no work is performed.

The reference points for the potentials are in the homogeneous field at the negatively charged metal plate and in the inhomogeneous field at infinite distance.

- Wikipedia: Article about "Electric field"
- Wikipedia: Article about "Potential energy"
- Wikipedia: Article about "Work (electrical)"

- Deutsche Version: Artikel über "Elektrische Felder II"