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4

You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use $\partial_x = \dfrac{d}{dx}$ for comfort, that being said, your equation can be written as: $$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...


3

You forgot to draw two exitting field lines. It's a hyperbolic singular point. It looks like this: Two in along the horizontal line, two out along the vertical line. You seem to have a mathematical problem with singular points. Sure... if you follow along the EXACT horizontal line, sure... you hit the point and don't know how to continue. But that's true ...


3

The flux lines around a point charge are spherically symmetric, so the total flux through a surface is proportional to the solid angle subtended by the surface. When you see questions like the one you cite they usually have some trick using symmetry. For example for the charge in the centre of the face the solid angle subtended by the whole cube is $2\pi$ ...


3

But I don't understand the mechanism of the force creation But the concept of electric charge and electric field is, by definition, the mechanism of the force creation - that humans have invented to model that which has been observed. Never forget that the observed is the metaphysically given. It is up to us, as beings possessing a rational faculty, ...


3

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

The magnetic force acting on a charged particle doesn't affect the particle's energy. Otherwise magnetic forces cannot do work. It's because the magnetic force equation is given by $$\vec{F}=q\vec{v}\times\vec{B}$$ where $q$ is the charge, $\vec{v}$ is the velocity and $\vec{B}$ is the magnetic flux density. So, it is clear that by virtue of the ...


2

On a local level, when you measure the electric field in matter (conductor or otherwise) there is a rapid variation both in time and space. Semiclassically, the electric field get's very large as you approach the nucleus, then drops off, the electrons are moving around the nucleus at very large speeds. So, yes, you are correct when you say there will be ...


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

Forget anything about a capacitor and just consider the resistance of the conducting liquid. Think of the liquid as made up of thin $dr$ concentric shells of radius $r$ and find the resistance of a shell in terms of the resistivity, radius and thickness. Then do the integration to find the resistance of all the liquid.


2

The electric field between the conductors is due to both sets of charges. however when finding a value for the electric field using Gauss's law it is only the charges inside the surface which are of interest and it is easier to choose the charge on the centre conductor and the red Gaussian surface $S_+$ which would be a cylider. You could find the ...


2

Potential refers to a particular point - or set of points which are "equipotential". So you can talk about the potential of one of the capacitor plates (because each is an equipotential surface) but not the potential of the capacitor (because when charged the $2$ plates are at different potentials). When talking about a capacitor, potential usually means ...


2

The assumption you have to make is that the zero of potential is infinity. The total work done by unit positive charge in bringing it from infinity to point $P$ is the potential at point $P$. $$-\int_\infty^r \frac{1}{4\pi\epsilon_0}\left(\frac{+q}{r^2}\right) dr -\int_\infty^{2r} \frac{1}{4\pi\epsilon_0}\left(\frac{-2q}{r^2}\right)dr = ...


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

Imagine that the electric field in the region of the neutral (null) point looks something like this with a Gaussian surface which is a cylinder shown in red. The exact shape does not matter. As drawn all looks well in that grey electric field lines enter at faces $A$ and $B$ and leave out of curved area $C$. One can imagine that all the sums work out as ...


2

He meant that the "energy" function $$ E_\textrm{art}(C) = - \frac{1}{2}CV^2 $$ is introduced solely for the purpose of getting the right result with the "principle of virtual work". It is not really EM energy in the usual sense as energy stored in the capacitor, available for use. If there is potential difference $V$, the latter energy $E$ is actually ...


2

If you ignore the coefficients, your function will be: f(x) = x / (x2 + r2)3/2 differentiating w.r.t x will give you: simple product rule f(x) = u(x)*v(x); u(x) = x; v(x) = 1 / (x2 + r2)3/2 f'(x) = u'(x)*v(x) + u(x)*v'(x) f'(x) = [1 / (x2 + r2)3/2] + [(-3/2)2x2 / (x2 + r2)5/2] to maximize/minimize you substitute f(x) = 0: 1 / (x2 + r2)3/2 = 3x2 / (x2 ...


2

$\nabla \cdot \vec E(r) = \dfrac {1}{r^2} \dfrac {d(r^2 E)}{dr} \ne \dfrac{dE}{dr}$ in spherical coordinates.


2

The RHS side of Gauss's Law, that is the charge enclosed should remain the same is indeed true. The apparent confusion if any, should be in the LHS of the equation, the integral of the 'dot product' of field and area vectors. Consider the diagram, Now, when we take the dot product of the field vector with the area vector in the initial case, the field and ...


2

A dipole has two parameters magnitude and orientation. If you change the orientation the nature of the force also changes (i.e from attractive to repulsive), so what you are saying is if you drop a body on suppose say North Pole and it is free falling towards earth then on the South Pole it should free fall away from earth which is not the case. ...


2

You can't have strong enough electric fields to tear the proton away from the nucleus but it is really a very subtle thing and the inability is just "by a little bit". The strongest electric field that may exist is given by the Schwinger limit. In $\hbar=c=1$ units, the field is $m_e^2 /q_e$. Once you reach this value, electron-positron pairs start to be ...


1

I have found your question and the diagram a little difficult to interpret. I have redrawn you diagram to show a charge of $+Q$ on the outer shell and a charge of $-Q$ at the centre together with two conducting shells shaded grey. What else the electric field inside the conductors is zero. If there was an electric field then the mobile charge carrier ...


1

The mistake you made was essentially transforming the force twice. In the frame in which the matter is stationary (I'll call this the primed frame), you correctly found: $F'=q\sigma'/2\epsilon_0$ and in the frame in which the matter is moving (unprimed) you correctly found: $F=q\sigma/2\epsilon_0 \gamma^2$ Since $\sigma'=\sigma/\gamma$ this is: ...


1

This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction?


1

You haven't seen these effects because your eyes are rather insensitive to wavelength changes. A moderately resolving spectrometer can detect these changes quite easily. It's called "Stark effect" and it can be observed in atomic spectroscopy: https://en.wikipedia.org/wiki/Stark_effect. For magnetic fields the analog is called "Zeeman effect": ...


1

What you have forgotten is that $u$ is not a potential rather it is a number with no units. I do not know how you derived your expression for $u$ but here is a way without a lot of the intermediate steps. The arrangement you have described is similar to that of two parallel line charges with separation $2a$ and charge per unit length $\pm \lambda$ as shown ...


1

In vacuum, any two point charges bearing electric charge of the same sign will solely interact, if they are pinned at a particular distance, via the Coulomb force that is in $\sim \frac{q_1q_2}{r^2}$ as you say so that they will always repel no matter the distance. Now, if you take in vacuum any two charged pieces of the same material (even at the ...


1

The electric field due to the outer cylinder has no contribution inside. One way to view it using Gauss's law, the other way is that if you took a slice from that cylinder, and considered a point inside it other than the center, you'll find a point producing electric field in the near side of the point (small charge, small distance) and a corresponding arc ...


1

The real problem here is the slack use of vectors, what you found was the magnitude of electric field but you need to represent it in vector form as the question demands that. Your textbook answer is wrong as well. Change your reference book. That being said, your calculations are correct.


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For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ...


1

Since the electron is negatively charged, the index finger should point up. The force is to the right, so the thumb points to the right. And then the middle finger comes out of the page. You said you have your thumb pointing to the left because the electron is negatively charged. But in that case you should consider the current to be to the bottom of the ...



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