And again:
For checking your understanding please do the following exercise:
Every day life teaches us that there are various charges and their effects. The image Abbildung 1 depicts a chargeable body that can be charged through charge separation between the sole and the floor. The movement of the foot generates a negative surplus charge in the body, which progressively spreads throughout the body. A current can flow even through the air if a pointed portion of the body (e.g., a finger) is brought into close proximity to a charge reservoir with no extra charges.
First, we shall define certain terms:
Only electrostatics is discussed in this chapter. For the time being, magnetic fields are thus excluded. Furthermore, electrodynamics is not covered in this chapter and is provided in further detail in subsequent chapters.
The concept of a field will now be briefly discussed in more detail.
Vector fields are defined as follows:
Place a negative charge $Q$ in the middle of the simulation and turn off the electric field. The latter is accomplished by using the hook on the right. The situation is now close to reality because a charge appears to have no effect at first glance.
A sample charge $q$ is placed near the existing charge $Q$ for impact analysis (in the simulation, the sample charge is called „sensors“). The charge $Q$ is observed to affect a force on the sample charge. At any point in space, the magnitude and direction of this force can be determined. In space, the force behaves similarly to gravity. A field serves to describe the condition space changed by the charge $Q$.
Abb. 2: setup for own experiments
Take a charge ($+1~{ \rm nC}$) and position it.
Measure the field across a sample charge (a sensor).
We had already considered the charge as the central quantity of electricity in block02 and recognized it as a multiple of the elementary charge. Now, we want to determine the electric field of charges. For this, a measurement of its magnitude and direction is now required. The Coulomb force between two charges $Q_1$ and $Q_2$ is:
\begin{align*} F_C = {{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2} \over {r^2}}} \end{align*}
The force on a (fictitious) sample charge $q$ is now considered to obtain a measure of the magnitude of the electric field.
\begin{align*} F_C &= {{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot q} \over {r^2}}} \\ &= \underbrace{{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1} \over {r^2}}}_\text{=independent of q} \cdot q \\ \end{align*}
As a result, the left part is a measure of the magnitude of the field, independent of the size of the sample charge $q$. Thus, the magnitude of the electric field is given by
$E = {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1} \over {r^2}} \quad$ with $[E]={{[F]}\over{[q]}}=1 ~{ \rm {N}\over{As}}=1 ~{ \rm {N\cdot m}\over{As \cdot m}} = 1 ~{ \rm {V \cdot A \cdot s}\over{As \cdot m}} = 1 ~{ \rm {V}\over{m}}$
The result is therefore \begin{align*} \boxed{F_C = E \cdot q} \end{align*}
The unit of $E$ is $\rm 1 {{N}\over{As}} = 1 {{V}\over{m}} $
Be aware that in English courses and literature $\vec{E} $ is simply referred to as the electric field, and the electric field strength is the magnitude $|\vec{E}|$. In German notation, the Elektrische Feldstärke refers to $\vec{E}$ (magnitude and direction), and the Elektrische Feld denotes the general presence of an electrostatic interaction (often without considering exact magnitude).
The direction of the electric field is switchable in Abbildung 2 via the „Electric Field“ option on the right.
In the case of the force, only the direction has been considered so far, e.g., direction towards the sample charge. For future explanations, it is important to include the cause and effect in the naming. This is done by giving the correct labeling of the subscript of the force. In Abbildung 3 (a) and (b), the convention is shown: A force $\vec{F}_{21}$ acts on charge $Q_2$ and is caused by charge $Q_1$. As a mnemonic, you can remember „tip-to-tail“ (first the effect, then the cause).
Furthermore, several forces on a charge can be superimposed, resulting in a single, equivalent force.
Strictly speaking, it must hold that $\varepsilon$ is constant in the structure. For example, the resultant force in Abbildung 3 Fig. (c) on $Q_3$ becomes equal to: $\vec{F_3}= \vec{F_{31}}+\vec{F_{32}}$.
Abbildung 3 Fig. (d) shows that for a charged surface, the force on a charge on top of this surface is always perpendicular to the surface itself.
Abb. 3: direction of coulomb force
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Now we want to see, whether we can derive the required energy to displace a charge in the electric field.
Since we know the force on a charge in an electrical field $\vec{E}$ (= Coulomb-Force $\vec{F}_C = q \cdot \vec{E} $), we can borrow some relationships from mechanics for the energy $\Delta W$:
\begin{align*} \Delta W = \int \vec{F} d\vec{r} = q \int \vec{E} d\vec{r} \end{align*}
Looks familiar? Maybe not on the first sight. But we already had defined the fraction of the energy difference per charge ${{\Delta W}\over{q}}$ as voltage $U$!
Therefore:
\begin{align*} \boxed{U = \int \vec{E} d\vec{r} } \end{align*}
We will apply this relationship in multiple of the upcoming blocks.
Given is the arrangement of electric charges in the picture on the right.
The following force effects result:
$F_{01}=-5 ~\rm{N}$
$F_{02}=-6 ~\rm{N}$
$F_{03}=+3 ~\rm{N}$
Calculate the magnitude of the resulting force.
The forces have to be resolved into coordinates. Here, it is recommended to use an orthogonal coordinate system ($x$ and $y$).
The coordinate system shall be in such a way, that the origin lies in $Q_0$, the x-axis is directed towards $Q_3$ and the y-axis is orthogonal to it.
For the resolution of the coordinates, it is necessary to get the angles $\alpha_{0n}$ of the forces with respect to the x-axis.
In the chosen coordinate system this leads to: $\alpha_{0n} = \arctan(\frac{\Delta y}{\Delta x})$
$\alpha_{01} = \arctan(\frac{3}{1})= 1.249 = 71.6°$
$\alpha_{02} = \arctan(\frac{4}{3})= 0.927 = 53.1°$
$\alpha_{03} = \arctan(\frac{0}{3})= 0= 0°$
Consequently, the resolved forces are:
\begin{align*} F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{x,0n} = F_{0n} \cdot \cos(\alpha_{0n}) \\ F_{x,0} &= (-5~\rm{N}) \cdot \cos(71.6°) + (-6~\rm{N}) \cdot \cos(53.1°) + (+3~\rm{N}) \cdot \cos(0°) \\ F_{x,0} &= -9.54 ~\rm{N} \\ \\ F_{y,0} &= F_{y,01} + F_{y,02} + F_{y,03} && | \quad \text{with } F_{y,0n} = F_{0n} \cdot \sin(\alpha_{0n}) \\ F_{y,0} &= (-5~\rm{N}) \cdot \sin(71.6°) + (-6~\rm{N}) \cdot \sin(53.1°) + (+3~\rm{N}) \cdot \sin(0°) \\ F_{y,0} &= -2.18 ~\rm{N} \\ \\ \end{align*}
Given is the arrangement of electric charges in the picture on the right.
The following force effects result:
$F_{01}=-5 ~\rm{N}$
$F_{02}=-6 ~\rm{N}$
$F_{03}=+3 ~\rm{N}$
Calculate the magnitude of the resulting force.
Given is the arrangement of electric charges in the picture on the right.
The following force effects result:
$F_{01}=+2 ~\rm{N}$
$F_{02}=-3 ~\rm{N}$
$F_{03}=+4 ~\rm{N}$
Calculate the magnitude of the resulting force.
Intro into electric field