The electrical potential

1 Work in an electric field

If a charge is located within an electric field, a force is exerted on it. If the charge is then moved within the field, work must be done. The gravitational field can be viewed analogously here, in that work must be done when a mass is lifted, i.e. the potential energy of the object is increased.

This section should be seen as a brief preview of Module 2 Energy and Power, in which electrical work is defined precisely. The sign of the work done depends on whether the work is done on the system under consideration (here: potential energy of the test charge \(Q\) in the electric field \(\vec {E}\))

or to the generator of this work. This script refers to the generator. If the generator has to perform work to increase the potential energy of the system, the work done is considered negative. If work is released by the system, i.e. the potential energy of the sample charge decreases, the work is positive.

Physically speaking, the work is the product of the force required to move the charge multiplied by the distance travelled. Using the knowledge from Section 4.1 that the electric field strength is the quotient of the force on a charge per charge (see equation ??), the work required to move the charge can be determined.

Below, we will calculate the work \(W\) released by the positive test charge \(Q\) from Figure 1 when it moves along the field lines in the direction of the negatively charged conductor plate.

The course of the electric field strength and direction of \(\vec {E}\) relative to the path \(x\) can be represented as shown in Figure 2.

\begin {equation*} W = \int _{d}^{0} \vec {F} \, d \vec {x} \end {equation*}

Since the direction of the electric field \(\vec {E}\) is opposite to the spatial direction \(x\), eliminating the vector arrows results in a negative sign in front of the integral. Substituting the force \(F\) into equation ?? yields:

\begin {equation*} W = - Q \int _{d}^{0} E \cdot \mathrm {d}x \end {equation*}

Solving the integral and inserting the limits of integration yields:

\begin {equation*} W = -Q \cdot E \, [x]_d^0 = -Q \cdot E (0-d) \end {equation*}

\begin {equation*} W = Q \cdot E \cdot d \end {equation*}

The difference between the work done or removed by a charge in an electric field is therefore:

\begin {equation} \Delta W = Q \cdot E \cdot \Delta x \end {equation}

For the sample charge shown in Figure 1, the aim is to determine how much energy is released on its way from the starting point to the negative plate.

The test charge is \(Q= 1 \mu \mathrm {C}\), the distance to the plate is 0.1 m, and the electric field is \(E= 200\) V/m.

The work \(W\) is then calculated as

\begin {equation*} W= Q \cdot E \cdot d = 1 \cdot 10^{-6} \mathrm {C} \cdot 200 \mathrm {V/m} \cdot 0,1 \mathrm {m} = 2 \cdot 10^{-5} \mathrm {J} \end {equation*}

Similar to an object in a gravitational field, the electric field of a charge causes energy. This accelerates the charge, and the potential energy is converted into kinetic Energy.

It is immediately apparent that the amount of potential energy stored in the electric field and usable by the test charge \(Q\) is linearly dependent on the location \(x\) of the test charge (see Figure 3). If the test charge already has kinetic energy in advance, or if the electric field extends beyond the section under consideration, the equation can be expanded by the additional term \(W_0\):

\begin {equation} W(x) = Q \cdot E \cdot x + W_0 \end {equation}

The transition from the potential energy stored in the electric field to the kinetic energy of the sample charge can be illustrated using the diagram in Figure 4.

For general description purposes, the technically usable field energy must be independent of the sample charge \(Q\). To this end, the potential energy is normalised to a working potential.

\begin {equation*} \frac {W(x)}{Q} = E \cdot x + \frac {W_0}{Q} \end {equation*}

2 Definition of electrical potential

The potential energy normalised to the charge Q under consideration is the electric potential \(\varphi \) of the electric field. This can be determined directly for every point in space, independently of other influencing factors. In contrast to the electric field, the potential field is a scalar field. This means that a unique electric potential can be assigned to every point in space independently of other influencing factors, but there is no direction for this quantity in space.

\begin {equation} \varphi = \frac {W}{Q} \end {equation} \begin {equation*} [\varphi ] = \text {V} \ (\text {Volt}) \end {equation*}

Figure 5 shows an electric homogeneous field in which the potential line on which the charge \(Q\) is currently located is plotted. As the distance from the positive plate increases in the \(x\) direction, the potential rises and can be calculated as follows:

\begin {equation} \varphi (x) = E \cdot x + \varphi _0 \label {eq:potential} \end {equation}

The gradient of the potential is determined by the electric field strength \(\vec {E}\). The electric potential \(\varphi \) and the electric field strength are related by the negative gradient of the potential. This relationship is illustrated in Figure 6.

The relationship can be expressed in a homogeneous field by the following equation:

\begin {equation} E = - \frac {\Delta \varphi }{\Delta x} \end {equation}

In inhomogeneous, three-dimensional fields, the general equation applies, in which the gradient of \(\varphi \) is calculated in every spatial direction:

\begin {equation} E = - \mathrm {grad} \,\varphi (x,y,z) = \left ( \frac {\partial \varphi }{\partial x}, \frac {\partial \varphi }{\partial y}, \frac {\partial \varphi }{\partial z} \right ) \end {equation}

In order to fully describe the electric potential at a point, as shown in equation 4, a reference potential \(\varphi _0\) must be selected as a necessary boundary condition, which is clearly defined. In principle, this potential can be selected arbitrarily, but as a rule, a reference electrode is selected for which the potential is set to zero. In the case of the setup shown in Figure 5, this would be:

\begin {equation*} \varphi _0 = \varphi (x=0) = 0 \end {equation*}

3 Connection between work and potential

In a homogeneous electric field, the potential \(\varphi \) is, as shown in formula 4, exclusively dependent on the distance \(x\) to the reference electrode. This means that all surfaces perpendicular to the field have a constant potential. Such surfaces are called equipotential surfaces, as shown in Figure 7 by the green lines. No work needs to be done to move a charge along such a surface.

The potential energy \(W_{12}\) of a test charge \(Q\) moved by the electric field from point \(x_1\) to \(x_2\) can be described by the potential difference between these points. The order of the indices of the work \(W_{12}\) describes the direction of movement.

If a test charge is moved from the equipotential surface \(\varphi (x_1)\) to \(\varphi (x_2)\) as shown in Figure 7, it absorbs the following energy:

\begin {equation*} W_{12} = Q \cdot (\varphi _1 - \varphi _2) \end {equation*}

With the choice of reference surface \(\varphi _2 = 0\) made in Figure 8, the result for the work supplied is:

\begin {equation*} W_{12} = Q \cdot (\varphi _1 - 0) = Q \cdot \varphi _1 \end {equation*}

The following thought experiment illustrates another key property of the electric field.

As shown in Figure 9, a test charge \(Q\) is moved in a closed circuit between points 1-4 in a homogeneous field. The partial energies required for this process are considered. Thus, when moving from point 1 to point 2, work of \(W_{12} = Q \cdot (\varphi _1 - \varphi _2)\) is performed, and from point 4 to point 1, energy \(W_{41} = Q \cdot (\varphi _4 - \varphi _1)\) is performed.

The energy required for a complete cycle is calculated by adding up the four partial energies:

\begin {equation*} \begin {split} W_\mathrm {ges} = Q \cdot (\varphi _1 - \varphi _2) + Q \cdot (\varphi _2 - \varphi _3) + Q \cdot (\varphi _3 - \varphi _4) + Q \cdot (\varphi _4 - \varphi _1) \end {split} \end {equation*}

Since the charge \(Q\) is identical in all terms, factoring out \(Q\) yields:

\begin {equation*} W_\mathrm {ges} = Q \cdot (\varphi _1 - \varphi _2 + \varphi _2 - \varphi _3 + \varphi _3 - \varphi _4 + \varphi _4 - \varphi _1) \end {equation*}

\begin {equation*} \rightarrow W_\mathrm {ges} = Q \cdot 0 = 0 \end {equation*}

No energy is required for a closed cycle of a test charge in a homogeneous field, as it depends solely on the starting and ending points of the movement, but not on the chosen path.

Consequently, the electric field is also referred to as a vortex-free source field. This means that the field lines do not form closed lines, but each have a starting point (on positive charges) and an end point (on negative charges).

This observation applies not only to homogeneous fields, but to all electrostatic fields.

Work in the radial field In the following example, the test charge \(Q_1\) is to be moved from a distance \(r_1\) to a distance \(r_2\) to the fixed charge Q.

\(Q\) is 1 C, \(Q_1\) is 1 mC. What work is required if \(r_1 = 1\) m and \(r_2 = 50\) cm?

\[ W_{12} = Q_1 \int _{r_1}^{r_2} \vec {E} \cdot d\vec {r} \] \[ W_{12} = Q_1 \int _{r_1}^{r_2} \left ( \frac {Q}{4 \pi \varepsilon _0 r^2} \right ) dr \] \[ W_{12} = Q_1 \cdot \frac {Q}{4 \pi \varepsilon _0} \int _{r_1}^{r_2} \frac {1}{r^2} dr \] The primitive function of \(\frac {1}{r^2}\) is \(-\frac {1}{r}\). \[ W_{12} = Q_1 \cdot \frac {Q}{4 \pi \varepsilon _0} \left ( -\frac {1}{r} \bigg |_{r_1}^{r_2} \right ) = Q_1 \cdot \frac {Q}{4 \pi \varepsilon _0} \left ( -\frac {1}{r_2} + \frac {1}{r_1} \right ) \] where \(\frac {Q}{4 \pi \varepsilon _0} \frac {1}{r}\) corresponds to the potential \(\varphi (r)\).

\[ W_{12} = Q_1 \cdot (\varphi (r_1) - \varphi (r_2)) \]

Inserting the given values: \[ W_{12} = (1 \cdot 10^{-3} \, \text {C}) \cdot \frac {1 \, \text {C}}{4 \pi \varepsilon _0} \left ( \frac {1}{1 \, \text {m}} - \frac {1}{0{,}5 \, \text {m}} \right ) \] \[ W_{12} = -8,98755 \times 10^{6} \, \text {J} \] Consequently, approximately 8.99 megajoules must be expended to move the charge \(Q_1\).

4 Definition of electrical voltage

The energy that can be utilised from the electric field for a technical process is given by the potential difference between the starting point and end point of the path. Due to its fundamental importance in electrical engineering, it is given a name and formula letter, and is referred to as electrical voltage \(U\):

\begin {equation} U_{12} = \varphi _1 - \varphi _2 \label {eq:spannung} \end {equation}

\begin {equation*} [U] = \text {V} \ (\text {Volt}) \end {equation*}

The order of the indices indicates the direction of the voltage arrow, analogous to the potential difference. In Figure 10, the voltage \(U_{12}\) describes the potential difference between the potentials \(\varphi _1\) and \(\varphi _2\). Due to the properties of equipotential surfaces, it is completely irrelevant which point of \(\varphi _1\) is started from and which point on \(\varphi _2\) is the destination. The path chosen between the points is also irrelevant for the voltage between the points.

The relationship between the electric field strength and the potential is already known from equation 4 and is given (for a reference potential of \(\varphi _0 = 0)\) by \(\varphi (x) = E \cdot x\).

Inserted into the definition of voltage (equation 7), this results in:

\begin {equation*} U = \varphi (x= d) - \varphi (x= 0) = E \cdot d - E \cdot 0 \end {equation*}

The electrical voltage \(U\) can be calculated in a homogeneous field as shown in Figure 11 as follows, provided that the distance \(d\) is parallel to the field lines:

\begin {equation} U = E \cdot d \end {equation}

Alternatively, the electric field strength can also be calculated from a given voltage \(U\) and a known distance \(d\):

\begin {equation} E = \frac {U}{d} \end {equation}

Figure 12 shows that the voltage is nothing more than the integrated field strength \(E\) over the distance \(d\).

While this integral can be determined in a homogeneous field by multiplying the electric field \(E\) and the distance \(d\), in inhomogeneous fields this is only possible by solving the integral:

\begin {equation} U_{12} =\int _{1}^{2} \vec {E} \cdot d \vec {s} \end {equation}

It should be noted here that the product \(\vec {E} \cdot \mathrm {d} \vec {s}\) consists of vectors. If they point in the same direction, the vector arrows can be eliminated; otherwise, the scalar product must be used:

\begin {equation*} \vec {E} \cdot \, \mathrm {d} \vec {s} = E \cdot \mathrm {d} s \cdot \mathrm {cos} (\alpha ) \end {equation*}