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19

You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


5

I actually agree with Ruslan's comment. You cannot say that the integral blows up when $\textbf{r} = \textbf{r}'$, with $\textbf{r}'$ spanning the integration domain where $\rho \neq 0$. The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function. To see this you ...


5

The E field due to each plate is $E/2$ and hence the total field between the plate is $E$. But a plate won't exert force on itself, so the E field experienced by a plate is $E/2$ only. Multiplying by charge gives you the force, hence $1/2 QE$.


5

The energy of the capacitor is $U= \frac{\epsilon_0}{2} S\,\mathrm d E^2$ where $S$ is the area of a plate. If we increase of $\Delta d$ the distance of, say, the right plate from the left one, keeping fixed the charge $Q$ on each plate, $E$ does not change and we find a variation of energy $$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = ...


4

The existing answers tell you why $F= \frac{1}{2} QE$ is right, but I think it's important to say why $F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}$ is wrong. Coulomb's law is not easily applicable here because the plates are not point charges. In particular, their sizes are not negligible (indeed, much larger) than the distance between them. It would ...


4

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...


3

As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge. Electric flux through a closed surface $\rm S$ is $$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.$$ Now, according to Divergence Theorem, \begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf ...


3

I would say yes, the electric field is well defined in a volume charge distribution. You raise some interesting points, that I think ultimately comes down to: Is the electric field consistent and well defined mathematically? What is the physical interpretation of the mathematics? Is a volume charge distribution physical and realistic? To answer the first ...


3

First, I think a little intuition could help. If you imagine a disk with a charge $Q$ get smaller, but all the while keeping the charge Q intact, shouldn't it geometrically approach a point with charge Q? So, in theory, we should expect the field to approach that due to a point charge. Now, intuition aside, let's go to the mathematics. While the factor ...


2

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 $\vec D$.


2

This is the direct consequence of gauss law.Consider a cuboidal gaussian surface as in the following figure.NOTE THAT GAUSSIAN SURFACE LIES JUST INSIDE THE PLATES THICKNESSSince electric field inside a conductor is zero, therefore no field lines pass through the surfaces enclosing charges, thus no flux through them. As for remaining surfaces, if we ignore ...


2

I think the key point here is that gravitational mass and inertial mass are not the same concept. The $m$ in Newton's second law, $F=ma$, is called inertial mass. One way to describe this might be the resistance and object has to be accelerated by a force. It takes more force to give a more massive object the same acceleration as that given to a less ...


2

Electromotive force (EMF), if generated, whenever the magnetic flux changes in time. In other words, it is proportional to its derivative viz. $$\mathscr E= -\frac{\mathrm d\Phi_\textrm{total}}{\mathrm dt}$$ where $\Phi_\textrm{total}= \textrm{total magnetic flux}= \Phi_\textrm{external} + \mathrm Li\;.$ So, it can be generated also in the moment when the ...


2

The electric potential of a point in space is defined by the mechanical energy that it takes to get a (positive) unit charge to that point. We usually define the potential of an infinitely distant point as zero, and then the movement of our test charge is from infinity to the point for which we want to measure the potential. In case of a capacitor the ...


2

When there are no external fields the charge must be distributed uniformly. If you have an external field and the sphere is made of conducting material then it will act as a Faraday cage and the charges will distribute themselves to cancel the field inside the sphere, leading to a nonuniform charge distribution on the surface with a 0 field inside the ...


2

The electric field induces some charge on the sphere. This charge on the sphere creates it's own electric field. The metal sphere will have a constant electric field throughout the surface as you said that the applied field is uniform. There is no charge inside. So there can't be any electric field inside the sphere. Otherwise it will violate Gauss's law ...


2

Good question! I watched Lewin's lectures a few years ago and distinctly remember this explanation as being unsatisfactory. It's one of the very few places where he "cheats" and skips a few steps. The missing step is the existence/uniqueness theorem. We know, from Lewin's argument, that the total charge on the inside surface is zero, and the boundary ...


2

A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...


2

We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...


2

I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...


2

You have misrepresented the citation in the book. The 5th edition page 757 discusses experiments with a hollow sphere and a solid sphere. The experiments verify that the exponent is 2 within experimental error.


2

The electric field at a point outside the volume charge distribution is well defined, certainly. What about the electric field at a point inside the charge distribution? Let $\vec{r'}$ denote a point that lies in the charge distribution and $\vec{r}$ denote a point where the electric field is to be determined. The problem is the determination of the electric ...


2

We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one from now on (in the SI unit system) for simplicity. Let ${\bf r}_0\in \mathbb{R}^3$ be a fixed position. The electric field in the $i$'th Cartesian direction at ${\bf r}_0$ is $$\tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} ...


1

Any non uniformity in charge accumulation is supposed to be distributed evenly since the plane conducts. Not true. The charge distribution will be such that there is no component of electric field parallel to the surface of the conductor - because that is what would generate a force on the charges, and cause them to be redistributed. Instead, what ...


1

Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question. Let $v$ be a small region of arbitrary shape, containing $P$ (defined by $\vec{r}$) in its interior. We consider the ...


1

No, there will be a net force because the charged object will induce a charge redistribution in the metal. See electrical induction.


1

In most cases (probably always), no. Suppose the charged object has a positive charge. Then it attracts the electrons in the metal close to itself, creating a positive charge on the region of the metal away from the charged object. The force of attraction on the electrons of the metal is more than the force of repulsion on the positive region of metal, ...


1

why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge $Q$ while the other plate has charge $-Q$ so there is no net electric charge stored. (2) If you increase ...


1

I assume that A is neutral to begin with. Then inside of A cannot be any charge by Gauss law. This means that the inside of A must neutral. Then you can take the volume between B and A. The boundary of that are the metal spheres A and B. On the whole boundary surface, there cannot be any electric field as that surface lies inside the metal. If there was any ...



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