In electromagnetism, a dielectric (or dielectric material) is an electrical insulator that can be polarized by an applied electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor, but instead only slightly shift from their average equilibrium positions, causing dielectric polarization. Because of dielectric polarization, positive charges are displaced in the direction of the field and negative charges shift in the direction opposite to the field (for example, if the field is moving parallel to the positive x axis, the negative charges will shift in the negative x direction). This creates an internal electric field that reduces the overall field within the dielectric itself. If a dielectric is composed of weakly bonded molecules, those molecules not only become polarized, but also reorient so that their symmetry axes align to the field.The study of dielectric properties concerns storage and dissipation of electric and magnetic energy in materials. Dielectrics are important for explaining various phenomena in electronics, optics, solid-state physics, and cell biophysics.
If I had two different materials between two oppositely charged plates, would the total dielectric strength just be an average of both dielectric constants with respect to each of their thicknesses? Or is it more complicated than that?
Material A is in full contact with one plate and material...
My thinking would be to do a work integral
Work = integral dWork
= integral delta V dq
= integral delta V 4(pi r^2) dr
The problem is, is this possible with a single integral?
Due to the offset cavity, the electric field E will not be constant at a given r.
We have a parallel plate capacitor with two different dielectrics
It seems to be the case that the potential difference on each half of the capacitor is the same.
Initially, the electric field was ##\vec{E_0}=\frac{2\sigma_+}{\epsilon_0}\hat{j}##.
If we were to insert a single dielectric...
Often in potential calculus problems, the uniqueness theorem of the solution of the Poisson problem with Dirichlet and Neumann boundary conditions is improperly "invoked," without bothering too much about making such an application rigorous, i.e., showing that indeed the problem we are solving...
Hello! I'm studying a Dielectrics subject in my master's degree which is a little distant from my Electronics area. The teacher asked for a report on the following graph.In this case, the class was about rupture in gaseous dielectrics. The teacher is a little scattered and presents the class in...
The model that he uses is a dielectric in which there is a spherical cavity with a dipole at its center. The dipole ##\vec{m}## has a component due to a permanent dipole and a component due to an induced dipole (because of polarization).
In order to obtain the dipole moment in the cavity, the...
A capacitor consisting of 2 square metal plates placed at a certain distance is connected to a potential difference generator V.
A slab of dielectric material is inserted into the space between the armatures.
By doing the calculation of the derivative of the electrostatic energy with respect...
So my idea was to separate the capacitor into two individual ones, one of length ##l - a## filled with a vacuum and one of length ##a## filled with the glass tube. The capacitances then are
$$
C_0 = \frac{2 \pi \varepsilon_0 (l-a)}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)}
$$
for the...
My attempt would be to calculate the electric fields of the vacuum and dielectric part seperately and then use superpositioning to obtain the full solution. However, I don't see an ##x##-dependency coming along that path. The assignment suggests that there must be one though. Unfortunately, this...
Question:
Solution first part:
Have I done it right?
I don't know how to begin with second part since the dielectric is non-lineair, and most formulas like $$
D=\epsilon E$$ and $$P= \epsilon_0 \xhi_e E$$, only apply for lineair dielectrics. What to do?
The correct answer to this problem is: ##\sigma = \varepsilon_0E\frac{\varepsilon-1}{\varepsilon}##
Here is my attempt to solve it, please tell me what is my mistake?
##E_{in} = E_{out} - E_{ind}##
##E_{ind} = E_{out} - E_{in}##
##E_{in} = \frac{E_{out}}{\varepsilon}##
##E_{ind} = E_{out} -...
I am confused about how the electric field changes in this problem - is E' = E/Ke=E/2? Is E = V/d a correct usage?
When I solve it this way, the answer is incorrect:
change in energy density = (1/2)ε(E'2- E2) = (1/2)ε(E2/4 - E2) = (1/2)ε(-3/4)(V/2d)2.
What am I doing wrong? Thanks.
The the electric field inside decreases due to the presence of a dielectric by a factor of dielectric constant K. Hence the force between the plates will decrease.
Is this right?
It is believed that Maxwell equations (together with other relations depending on the materials) are sufficient to account for any electromagnetic macroscopic effect.
The problem is that, for a Maxwell equation to hold, it must at least be defined.
Consider for example the case of two...
Hello,
for my PhD, I've been studying an influence of a gain medium on spectral linewidth of light inside a fiber cavity. I've encountered a formula in one paper to which I don't how to get to (see screenshot), it's the formula (A3).
On the left hand side there is electric flux density, the...
I tried to solve it for some time and then looked at the solution manual, which got me completely lost. Those are the first lines of the solution :
I'm not so sure how equation 4.39:
makes him conclude that the same relation holds for dipole moments. My second concern is that I'm not sure how...
I understand that the particle will be polarised according to its dielectric constant and the electric field across the capacitor.
However, since it is similar to an insulator and electrons do not move in and out of the particle easily, the particle will not be charged.
How then will...
I have found the total dipole moment of for this problem but am having trouble finding the electric field.
I believe my electric field when r>2R ( I mistakenly wrote it as r<2R on my work, but it is the E with a coefficient of 2/3) is correct as it fits the equation:
.
I don't believe this...
I know that inside region 1, the D-field is zero as it is a conducting sphere, the E-field must be zero. It makes sense that in region 2 (inside the dielectric) there is a D-field.
My question is, is there a D-field outside the dielectric material (r>R)? Obviously there will be an E-field, but...
Since there is no free charge ##\int_S \vec{D} \cdot d\vec{a} = 0## and
##\rho_f = 0##
##\sigma_f = 0##
##\vec{nabla} \cdot \vec{P} = 0## since P is a constant
##\rho_b = - \vec{nabla} \cdot \vec{P} = 0##
For a simple surface we can find the boundary conditions for ##\vec{E}## using a Gauss'...
The dielectric strength of air (ie the maximum electric field that the material can withstand under ideal conditions without undergoing electrical breakdown and becoming electrically conductive) is 3 000 kV ( https://en.wikipedia.org/wiki/Dielectric_strength#Break_down_field_strength ).
In...
I considered the capacitor as two capacitors in parallel, so the total capacitance is ##C=C_1+C_2=\frac{\varepsilon_0\varepsilon_1 (A/2)}{d}+\frac{\varepsilon_0\varepsilon_2 (A/2)}{d}=\frac{\varepsilon_0 A}{2d}(\varepsilon_1+\varepsilon_2).##
Since the parallel component of the electric field...
My guess is that while the voltage between the two plates is lower when a dielectric is present, the maximum voltage that the capacitor can hold will actually increase because the maximum strength electric field generated by the charges on the two plates will be higher due to the opposing...
Hello! I am a junior undergraduate physics major and I am very confused on how to visualize things in my electrodynamics class. Specifically, I am having issues with dielectrics and spheres with constant potentials etc. I usually notice that I am lost in a class when I can no longer draw out a...
hi guys
our professor asked us to confirm the units of volume charge density ρ and also the surface charge density σ of a dielectric material given by
$$
\rho = \frac{-1}{4\pi k} \vec{E}\cdot\;grad(k)
$$
$$
\sigma= \frac{-(k-1)}{4\pi} \vec{E_{1}}\cdot\;\vec{n}
$$
I am somehow confused about the...
I am just learning about e-field simulations and I came across two different types of dielectric constants: optical and static. I'm unsure which to use and in which cases. I would like to simulate e-field intensity to help ensure I'm always below the dielectrics breakdown strength.
Regarding the electrical permittivity of the metal in a high frequency regime, I cannot find research material related to the lead dielectric function (PD). I can't get the matatrial as values, I'll let you comment on that. I know that Pd can inhibit the amount of gamma rays in the x-ray case...
The net Electric field(inside the dielectric):
$$E_{net} = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{q}{r^2}$$
$$\vec E_{net} = \vec E_{applied} - \vec p$$
where p is the polarization vector.
let charge ##q_{-}## be present on the inner surface of dielectric and ##q_{+}## on the outer...
Hi there,
if a dielectric (capacitor) is described with a constant permittivit eps (or C) and loss-tangent DF, how much energy ist lost when charging the capacitor by 1V?
For example: C=1, DF=0.1.
When charging from 0 to 1V, the lost energy (in J) is ...?
When charging from 1V to 2V, the lost...
In the 7th edition of the book "Elements of Electromagnetics by Matthew N. O. Sadiku"
On page 190 the author goes on to say:
"We now consider the case in which the dielectric region contains free charge.
If ##\rho_v## is the volume density of free charge, the total volume charge density...
I pretend to use the ecuation twice, once for the interior and another for the vaccum, so if I use the cilindrical coordinates for \nabla_t^2 it results in two Bessel equations, one for the interior and another fot the vaccum.
In the vaccum, the fields should experiment a exponential decay, in...
Hey guys! I'm having trouble with the solution that I arrived at.
Through boundary conditions I'm able to determine ##\vec{D}## as $$\vec{D}=-\frac{4Q}{R_0^2}\hat{e_z}$$ (In CGS units)
Trough that I'm able to get the electric field as $$\vec{E}=-\frac{1}{\epsilon(r)}\frac{4Q}{R_0^2}\hat{e_z}$$...
Hi,
On slide 9 of this presentation: http://www.globalcommhost.com/rogers/acs/techsupporthub/en/docs/MWJ_webinar_June20_2017_JC_microstrip_coplanar_stripline_final.pdf it states the signal wavelength can be changed with dielectric constant Dk.
As far as I understand the wavelength and the...
First, I think that I need to calculate the capacitance. It is ## C=\epsilon_0*\frac{l*w}{d}-x*\frac{w*\epsilon_0}{d}+\epsilon*\frac{x*w}{d} ##. After that I should calculate the potential energy. It is ##U=\frac{1}{2}*C*V^2 ##. After that I should take its gradient to get the force. So ##\vec F...
I am interested in the physics of ground penetrating radar.
1) Does anyone know where I can find a derivation of the formula in the attached jpg for the energy reflected? K=the dialectic constant.
2) An intuitive explanation of why the dialectic constant is important in determining the energy...
I am not able to intuitively understand the reasoning behind why the presence of dielectric between oppositely charged plates, let's say, reduces the force of attraction between the plates. I understand to some extent that electric field lines prefer to flow through dielectric (or insulator)...
I use the following equations to understand this question/answer.
First, C = k(ε*Area)/distance = Q/V = Q/ (E*distance)
As a slab of glass is added, k increases and thus E decreases.
F=QE, as E decreases, force decreases as well. How does this relate to the 'force attracts the glass into the...
I tried approaching this by finding the tangential and normal electric fields. Is this the correct approach? I've attached a drawing of the surface provided.
##\oint_S E \cdot dl=0##
##E_{tan1}\Delta x-E_{tan2}\Delta x=0##
We know that
##E_{tan1}=E_{tan2}
Next, we can find the normal...
Lets go through the example problem until we get to the part I don't understand. Figure 25-17 can be used as a reference to all questions. From part (a) to part (b) we eventually find the charge q on one plate (and by default the charge -q on the other). No problem there. The battery is then...
First when it is connected to the battery, the capacitors start accumulating charges such that the potential difference equals that of the battery. Then the current stops flowing.
##Q_1 = CV##
##Q_2 = nCV##
Where 1 and 2 represent the capacitor with capacitance C and nC respectively
Then, when...