Im trying to calculate the dielectric constant of a dielectric sample placed inside a rectangular waveguide over a range of frequenices. I used WR90 ( X band ).Using the s parameter data (S11 and S21), I used the Nicholson-Ross-Weir method to calculate the material parameters. But Im not getting a constant value(expected), rather Im getting wide variation in the dielectric constant. Given paper shows what i simulated in CST.
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Im trying to find the material paramters of a dielectric sample using a waveguide(X band) employing the Transmission/Reflection technique. While performing simulation in CST, how to properly orient the sample inside? Is it necessary that the waveguide is flared?
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This is a question from the second chapter (Problem 2.55) of Griffiths 'Introduction to Electrodynamics' fourth edition. The question says that if in a region, the electric field $E$ exists such that, $E_x = ax$ where $a$ is a constant, $E_y=0$ and $E_z=0$, what is the charge density. The charge density is given by the differential Gauss law, where, $\nabla.E = \dfrac{\rho}{\epsilon_0}$. This gives $\rho=\epsilon_0a$. Since $a$ is a constant, the charge density is uniform everywhere in the region. Now Griffiths further asks how could this uniform charge density produce an electric field that is parallel to just $X$ axis?  My intuitive answer is if there is an infinite slab like uniform distribution of charge where the slab has infinite length and breadth but a finite thickness, if the thickness is along the $X$ axis, the electric field will be along $X$ axis given as $E_x = ax$, $E_y=0$ and $E_z=0$.  However Griffiths himself answers this question in the instructors solution manual as given below.  While I also agree to Griffiths on a broader picture, I want to know if my intuitive solution of an infinite slab of finite thickness wrong?
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I ran across an interesting integral and I am wondering if anyone knows where I may find its derivation or proof.  I looked through the site. If it is here and I overlooked it, I am sorry.  $$\displaystyle\frac{1}{\int_{-\infty}^{\infty}\frac{1}{(e^{x}-x)^{2}+{\pi}^{2}}dx}-1=W(1)=\Omega$$  $W(1)=\Omega$ is often referred to as the Omega Constant. Which is the solution to   $xe^{x}=1$.  Which is $x\approx .567$  Thanks much.   EDIT:  Sorry, I had the integral written incorrectly. Thanks for the catch.  I had also seen this:  $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\approx .638$  EDIT:  I do not what is wrong, but I am trying to respond, but can not. All the buttons are unresponsive but this one. I have been trying to leave a greenie and add a comment, but neither will respond. I just wanted you to know this before you thought I was an ingrate. Thank you. That is an interesting site.
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I am trying to solve an integral in $k$-space where I want to calculate the ratio of the average energy of an arbitrary electronic band ($E(k)=k^n)$ to the square of average group velocity of electrons($v_g   = k^{n-1})$ approximately in the form of this expression given by $$L=\frac{\frac{\sum_{\mathbf{k}}k^{n}}{\sum_{\mathbf{k}}1}}{\left(\frac{\sum_{\mathbf{k}}k^{n-1}}{\sum_{\mathbf{k}}1}\right)^2}\;.$$  Now, I want to do this for $d$-dimensions, so I convert this sum to integral.  $$L=\frac{\dfrac{\int_{0}^{k_f}k^{n} d^d k}{\int_{0}^{k_f}1d^d k}}{\left(\dfrac{\int_{0}^{k_f}k^{n-1}d^d k}{\int_{0}^{k_f}1d^d k}\right)^2}$$ where $d^dk  = \dfrac{d \pi^{d/2}}{\Gamma(\frac{d}{2}+1)}k^{d-1}dk$ is the volume element in $d$-dimensional space.  I want to get a simple expression for $L$ and how it depends with $k_f,n$ and $d$. any physical insight into this integral could also be helpful.
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What is the formula for capacitance of two non parallel plates at an angle with each other?If the plates were parallel then the value can be calculated as (PermittivityX area of one plate)/distance between them.But what happens in case the plates are tilted at an angle?The question came to mind while trying to understand electrostatic separator.What would be the derivation of the formula for capacitance of two non parallel plates placed at an angle? I did get a method from  Equation 6 from the above link above helps but it is independent of the length of the plates which doesn't seem likely.
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I was studying a  problem where optical light with a wavelength of 500&nbsp;nm travels through a gas with refractive index $n$.  If we look at the equations for wave motion and index of refraction  $$c=\lambda_0\nu\quad\text{(in vacuum)}$$  $$v = \lambda\nu\quad\text{(in medium)}$$  $$n = c/v$$  we see that, if the frequency is constant, the wavelength decreases in the medium compared to vacuum. Is this a consistent property at all frequencies and for all mediums with refractive index real and greater than 1?  Are there dielectrics which change the frequency (still for n > 1), and is there an example of that?
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I know that the  is the leftover radiation from the "".   However, at every instant the surface is changing (at the rate of flow of time). So how constant is the CMBR?
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I'm not sure how to go about this. Does it involve using MVT?  I got as far as saying $e = \frac{e^x - e^y}{x-y}$.
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Below we have the Fresnel Equation at an interface for S and P polarisation: \begin{align}  r^s &amp;= \frac{n_0 \cos{\theta_0}-n_1\cos{\theta_1}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}} \quad\quad t^s = \frac{2n_1\cos{\theta_0}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}}\\  r^p&amp;= \frac{n_1 \cos{\theta_0}-n_0\cos{\theta_1}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}}\quad\quad t^p = \frac{2n_1\cos{\theta_0}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}}  \end{align} When light approaches the interface at normal incidence these equations simplify to: \begin{align}  r^s &amp;= \frac{n_0 -n_1}{n_0 + n_1} \quad\quad t^s = \frac{2n_1}{n_0 + n_1}\\  r^p&amp;= \frac{n_1 -n_0}{n_1 + n_0}\quad\quad t^p = \frac{2n_1}{n_1 + n_0}  \end{align}  When we apply this to an example where we have an aluminium mirror with refractive index, for example, of 0.93878 + i6.4195.   This would lead for S polarisation to a reflection coefficient -0.9138 - 0.2855i, leading to reflectivity of 95% but a phase shift of -0.90$\pi$. However, for P polarisation a reflection coefficient  0.9138 + 0.2855i is found, leading to the same reflectivity but a phase shift of 0.09$\pi$.  Since we come in under normal incidence, why would the orientation of the polarisation matter? The polarisation S and P are both parallel with the surface, so this would mean if I rotate the mirror I would change the phase of the reflected beam? How does this make sense? What am I missing?   Edit: this also the case for dielectric matter (i.e. real $n$)
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When considering the Doppler shift, the 'canonical equation' is  $$f=\frac{c+vr}{c+vs}f_0$$  However, this equation seems to run into trouble in the following situation: A light source inside water is moving at a speed $v$ towards a receiver outside the water. Can we modify the above equation to deal with a transition between two media? Or is there a completely different formula that applies here?
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I have just finished learning the basics of magnetism, and it should be noted that I am not very familiar with Maxwell's equations.  Note:    In the question, when I say , I am referring to the equation without Maxwell's correction. Also, when I say , I am referring to the equation: $\mathrm dB= (\mu_0/4\pi)(I)(\mathrm dL~ X~\hat r)/r^2$     Consider an infinitely long straight wire, carrying a time varying current I(t) such that dI(t)/dt is non-zero. Also consider a point P which is at a distance r from the wire. Using Biot Savart Law, we find out that the magnetic field is $\mu_0\cdot I(t)/2\pi \cdot r$, at any instant t.   Now, I have read that Ampere's Circuital law is NOT valid for cases in which the currents are time varying. However, if we consider an Amperian loop along a circle of radius r and centre at the perpendicular from P to the wire, using symmetry arguments, we obtain the same value of field: $\mu_0 I(t)/2\pi\cdot r$. Since Ampere's law is invalid for such a current, the expression mentioned for the magnetic field must be incorrect.  So, can Biot Savart Law also NOT be used for time varying currents? Also, just out of curiosity, what would be the actual value of the magnetic field at time t?  My book (Halliday and Resnick) derives the equation for the magnetic field created due to a moving point charge. However, after the derivation, it states that the result obtained is not really valid, since "a point charge cannot be assumed as a steady current by any stretch of imagination". This makes me believe that even Biot-Savart Law is only true for non time varying currents. Am I right or wrong?
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I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.         &ndash; Schematic created using   Let \$ H(s) = H_1(s)H_2(s) \$ where \$ H_1(s) \$ and \$ H_2(s) \$ are the transfer functions for each separate filter stage.  Then \$ |H(s)| = |H_1(s)||H_2(s)| \$  Knowing the magnitude of a passive low pass filter,   $$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$  Then trying to find the cutoff frequency:  $$ \left(\dfrac1{\sqrt{2}}\right)^2  = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$ $$ 2  = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)} $$ $$ 4  = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1) $$ $$ 4  = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1 $$  And I'm stuck. Research on the web tells me \$ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} \$, but I can't find why? Can anybody show my the derivation to find this?
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I am currently designing by CP microstrip antenna using CST. The picture is shown below. However, when I view the axial ratio plot, also shown below it is way above 3dB at 90 degrees. the patch is 31,95 by 31.95mm.  The resonate frequency output is correct which is 915MHz. I have tried varying multiple variables such as the length of the patch, length of the main feed, the distance between the branched feed line and the patch (quater-wavelength feed line). But the axial ratio does not change much. How can I alter the axial ratio to be below 3dB but keep the resonate frequency at 915MHz?
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I'm trying to understand how oscillators work and am currently working with an RC phase shift oscillator:    $$R_1, R_2, R_3 = 200 \Omega$$  $$C_1, C_2, C_3 = 4.7\text{nF}$$  one of the issues I'm having is that one source states that \$R_{\text{fb}} = 29R\$ for sustained oscillation yet when I set the resistor to \$5.8\text{k}\Omega\$ I get a flat line in nV.   What am I missing out?
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What was the improvement that Maxwell did to the  and why?  I understand that he combined the main equations so that you could get a wave equation for the vectors of $\bf E$ and $\bf B$ but is there anything else?
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Is it possible to derive Ohm's law (perhaps in some appropriate limit) from Maxwell's Equations?
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I understand that in a Frequency Response experiment dealing with an RLC circuit, the graph of Current against Frequency is supposed to be symmetrical about the resonant frequency theoretically.  However, experimentally it is not the case.  Could anyone explain why this happens?