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Assume that the voltage across an isolated capacitor with air ($\approx$ vacuum) between the plates of separation $d$ is $V_{\text{air}}$ and the charge stored on the capacitor is $Q$. $Q = C_{\text{air}} V_{\text{air}}$ The electric field between the plates is $E_{\text{air}} = \dfrac {V_{\text{air}}} {d} $. Now put a dielectric of relative ...


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Dielectric implies nonmagnetic, so the best example of an insulator that isn't a dielectric I can think of is ferrite crystals like yttrium iron garnet, that show a magnetic response while having low or zero conductivity. There are also ferroelectric materials that show a permanent polarization in response to an electric field, and these aren't considered ...


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In my understanding, the free charge is any charged particle that is not being restrained in the boundary, while the bound charge is in the boundary.It does not matter whether the material you currently discuss is a dielectric or a conductor.


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Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ...


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I assume you are talking about linear materials with dielectric constant greater than $1$. Say you have a free charge distribution $\rho$ in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite ...


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(For completely filled capacitors) Q = CV So, C = Q/V So, C is charge stored per unit Potential Difference applied. Now, V = Ed ,where d is distance between plates. $E = \dfrac{V}{d}$ Case 1) When you apply a constant V of 1V to capacitor E across capacitor is $ \dfrac{1V}{d}$ which is constant independent of capacitance of capacitor or dielectric b/w ...


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a) Capacitance C increases b) Charge Q remains unchanged c) if charge Q is constant while C increases, that means voltage V decreases (C=Q/V). d) U decreases ( U=Q^2/2C)


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If you find my other answer too long, this is the short answer. Step 1: Just ignore the word "dielectric" and consider it as a uniformly charged sphere (that you may have already solved before). Use Gauss's law to find the E field inside the outside. Step 2: For the field inside, replace $\epsilon_0$ by $K \epsilon_0$.


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Inside matter, in addition to free charges $\rho_f$ (which I suppose is what your $\rho$ means), there are bound charges $$\rho_b=-\nabla \cdot {\bf P}$$ as well. One way to solve the problem is to stick to the two equations $$\nabla \cdot {\bf E} = \rho/\epsilon_0$$ $$\nabla \times {\bf E} = {\bf 0}$$ But then you have to pay attention here that ...


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Dielectric constant $K$ is actually the same thing as relative permittivity, and it increases the overall permittivity $\epsilon$. So in general, whenever you see the permittivity of free space $\epsilon_0$ in an equation, if you're dealing with a dielectric, you can multiply it by the dielectric constant and see how the equation changes. For example, since ...


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A displacement field is produced by the polarization of a dielectric. A displacement current is produced by a time-varying electric field. The two concepts are completely different. A displacement field does not cause displacement current, and a displacement current is not affected by displacement field. There is no displacement current in a dielectric ...


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If the dielectric has permittivity $\epsilon = \epsilon_r \epsilon_o$, where $\epsilon_r$ is the relative permittivity or dielectric constant of the dielectric and $\epsilon_o$ is the permittivity of free space, then $\iint_S \epsilon \vec E \cdot d\vec A = Q$ is the form of Gauss's law to used. $\epsilon \vec E$ is called the displacement $\vec D$.


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(a) Of course after redistribution of charges among plates $A$ and $D$, an excess negative charge of $60\mu C$ will remain on the wire connecting plates $A$ and $D$. But remaining charge of D is not $30\mu C$. You have $V_C-V_D=V_B-V_A$ and $12\mu F (V_C-V_D )+ 18\mu F (V_B-V_A)=60\mu C$. So charge of plate $D$ is $-\frac{18\mu F}{ 12\mu F+ 18\mu F } ...


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(a)You need to have the potential difference across both capacitors to be the same so as their capacitances are different so must be the charges on their plates. So think about $V= \frac QC$ to decide how the net charges are distributed. (b) $C_2$ can be thought of a combination of two capacitors in series and that combination then being in parallel with ...


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You just need the phasor representation of a time derivative. $P e^{i \omega t} \rightarrow \vec{P}$ $\partial_{t}(P e^{i \omega t}) \rightarrow i \omega \tilde{P}$ Now, since as you said $J=\partial_{t}P$, we have: $\tilde{J}=i\omega \chi \tilde{E}$ $\langle \vec{E} \cdot \vec{J} \rangle = \frac{1}{2}\text{Re}(i \omega (\chi' + ...


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how the direction of the force on slab in both situation differs? Recall that the energy stored in a capacitor is given by $$W = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$$ where $V$ is the voltage across the capacitor and $Q$ is the magnitude of the electric charge on either plate. For the case that a constant voltage source (e.g., battery) is ...


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Consider shape, capacitance, size , max. voltage/current it can withstand. Try to look at some capacitor, may be from your house fan. (They generally burn after some time.) You will find your answer. Economically, CuriousOne answers correctly.


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Good and important question, altough one does not 'simply substitute'. The short answer is that the vacuum values are for microscopic theory of light in vacuum, and any other values are for macroscopic theory, which handles light-matter interaction phenomenologically at much larger scales than atomic separation. Here is the very long answer: The ...



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