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*}
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: $\vec{F_3}= \vec{F_{31}}+\vec{F_{32}}$.
Abb. 3: direction of coulomb force
.
In previous chapters, only single charges (e.g., $Q_1$, $Q_2$) were considered.
$\rho_l = {{Q}\over{l}}$
or, in the case of different charge densities on subsections:
$\rho_l = {{\Delta Q}\over{\Delta l}} \rightarrow \rho_l(l)={{\rm d}\over{{\rm d}l}} Q(l)$
$\rho_A = {{Q}\over{A}}$
or if there are different charge densities on partial surfaces:
$\rho_A = {{\Delta Q}\over{\Delta A}} \rightarrow \rho_A(A) ={{\rm d}\over{{\rm d}A}} Q(A)={{\rm d}\over{{\rm d}x}}{{\rm d}\over{{\rm d}y}} Q(A)$
$\rho_V = {{Q}\over{V}}$
or for different charge density in partial volumes:
$\rho_V = {{\Delta Q}\over{\Delta V}} \rightarrow \rho_V(V) ={{\rm d}\over{{\rm d}V}} Q(V)={{\rm d}\over{{\rm d}x}}{{\rm d}\over{{\rm d}y}}{{\rm d}\over{{\rm d}z}} Q(V)$
Electric field lines result from the (fictitious) path of a sample charge. Thus, also electric field lines of several charges can be determined. However, these also result from a superposition of the individual effects - i.e., electric field - at a measuring point $P$.
The superposition is sketched in Abbildung 4: Two charges $Q_1$ and $Q_2$ act on the test charge $q$ with the forces $F_1$ and $F_2$. Depending on the positions and charges, the forces vary, and so does the resulting force. The simulation also shows a single field line.
For a full picture of the field lines between charges, one has to start with a single charge. The in- and outgoing lines on this charge are drawn equidistant from the charge. This is also true for the situation with multiple charges. However, there, the lines are not necessarily run radially anymore. The test charge is influenced by all the single charges, and therefore, the field lines can get bent.
In Abbildung 6 the field lines are shown. The additional „equipotential lines“ will be discussed later and can be deactivated by clearing the checkmark Show Equipotentials
.
Try the following in the simulation:
Add
» Add Point Charge
.delete
There are two different types of fields:
In homogeneous fields, magnitude and direction are constant throughout the field range. This field form is idealized to exist within plate capacitors. e.g., in the plate capacitor (Abbildung 7), or the vicinity of widely extended bodies.
For inhomogeneous fields, the magnitude and/or direction of the electric field changes from place to place. This is the rule in real systems, even the field of a point charge is inhomogeneous (Abbildung 8).
Sketch the field line plot for the charge configurations given in Abbildung 9.
Note:
You can prove your result with the simulation Abbildung 4.
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.
Given is an arrangement of electric charges located in a vacuum (see picture on the right).
The charges have the following values:
$Q_1=7 ~\rm{µC}$ (point charge)
$Q_2=5 ~\rm{µC}$ (point charge)
$Q_3=0 ~\rm{C}$ (infinitely extended surface charge)
$\varepsilon_0=8.854\cdot 10^{-12} ~\rm{F/m}$ , $\varepsilon_r=1$
1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.
2. is this force attractive or repulsive?
Now let $Q_2=0$ and the surface charge $Q_3$ be designed in such a way that a homogeneous electric field with $E_3=100 ~\rm{kV/m}$ results.
What force (magnitude) now results on $Q_1$?
Given is an arrangement of electric charges located in a vacuum (see picture on the right).
The charges have the following values:
$Q_1=5 ~\rm{µC}$ (point charge)
$Q_2=-10 ~\rm{µC}$ (point charge)
$Q_3= 0 ~\rm{C}$ (infinitely extended surface charge)
$\varepsilon_0=8.854\cdot 10^{-12} ~\rm{F/m}$ , $\varepsilon_r=1$
1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.
2. is this force attractive or repulsive?
Now let $Q_2=0$ and the surface charge $Q_3$ be designed in such a way that a homogeneous electric field with $E_3=500 ~\rm{kV/m}$ results.
What force (magnitude) now results on $Q_1$?
Given is an arrangement of electric charges located in a vacuum (see picture on the right).
The charges have the following values:
$Q_1= 2 ~\rm{µC}$ (point charge)
$Q_2=-4 ~\rm{µC}$ (point charge)
$Q_3= 0 ~\rm{C}$ (infinitely extended surface charge)
$\varepsilon_0=8.854\cdot 10^{-12} ~\rm{F/m}$ , $\varepsilon_r=1$
1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.
2. is this force attractive or repulsive?
Now let $Q_2=0$ and the surface charge $Q_3$ be designed in such a way that a homogeneous electric field with $E_3=100 ~\rm{kV/m}$ results.
What force (magnitude) now results on $Q_1$?
Intro into electric field
Field lines of various extended charged objects