# Tag Info

0

One way to think of it is to imagine a point source pushing out water. If you put some imaginary surface around the source, the total amount of water passing through that surface is the same as the amount of water being pushed out by the source. It doesn't matter what the shape of this surface is, the total amount of water passing through it will be the same ...

0

If I'm understanding your question correctly, it's because we design the Gaussian surface that way. Here's what I mean by that: Gauss's law works for any arbitrary closed surface, regardless of its shape. The total flux through the surface must be proportional to the enclosed charge. For most surfaces you could invent (and you really are free to invent ...

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.

0

Alright, one issue I noticed is the incorrect Gauss' Law expression. It is actually: $$\oint {\vec E}\cdot \mathrm{d}{\vec A} = \frac{q}{\epsilon_0}$$ So only the perpendicular component of the vectors through each face surface matters. Since the electric field vectors for all the voltages you've given across each surface are acting upwards, only the ...

0

Determine the surface element of the cube, 2x2y2z or 8xyz, then use the integral version of Gauss's Law.

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 ...

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 ...

1

For a charge distribution $\rho({\bf r'})$, the electric field at ${\bf r}$ is $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where ${\bf R}={\bf r}-{\bf r'}$. I think the OP's claim is at positions where $\rho \ne 0$, the above integral is infinite because the integrand blows up at ${\bf R}={\bf 0}$. I think ...

1

Your thinking is correct. At least, if the sphere is made up of small point charges, then the field will be infinite as you approach them. There is a point you are missing when you say that this contradicts Gauss' law: Gauss' law only gives you the flux of the field. To get the field of the sphere from it in the textbook-way, you have to use symmetry. You ...

1

We can assume w.l.o.g. that the electric potential $$\left. \Phi(r,\theta,\varphi) \right|_{r<r_0}~=~0$$ vanishes in the interior. As already argued in Daniel Mahler's answer, a surface charge distribution $\rho$ with support at $r=r_0>0$ is far from unique. In fact, the reader may check that any electric potential of the form $$... 0 If you find my other answer too long, this is the short answer. Step 1: Just ignore the word "dielectric" and consider it as a uniformly charged sphere (that you may have already solved before). Use Gauss's law to find the E field inside the outside. Step 2: For the field inside, replace \epsilon_0 by K \epsilon_0. 1 Inside matter, in addition to free charges \rho_f (which I suppose is what your \rho means), there are bound charges$$\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 ...

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 ...

1

Yes, this is guaranteed by the uniqueness theorem for Poisson's Equation, and in fact is more general than spherically charged shells. However, as another answer indicates, that if there are other charges present elsewhere, the charge on the sphere will shift to cancel off the field interior of the conductor.

0

"The divergence of any electric field created by any surface charge distribution would be zero." No it's not. Consider a charged conducting sphere with uniform surface charge density and a Gaussian sphere of radius greater than the original one. The electric field is diverging through the surface of the Gaussian sphere. So the divergence cannot be zero. ...

0

Well, your statement is incomplete. Electrostatic field is zero inside the MATERIAL of the conductor. It may have any other value inside( inside does not include the material of the conductor) or outside the conductor. It is zero inside the MATERIAL of the conductor, because since charges are free to move, they generate a electric field due to polarised ...

0

All the points on the inner surface of the hollow conductor have the same potential. Therefore, if the charge density $\rho$ inside is zero everywhere, the electrostatic field will be zero. In your case the filament carries a charge $Q$, so the charge density is not zero everywhere.

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

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 $\vec D$.

0

The fact is the when you add the charge $-q$ the electric field changes. In particular it has no more spherical symmetry and, consequently, you cannot compute the flux through the gaussian surface in a simple manner, as you did when you had only one charge. Thus, in this case, Gauss theorem is, of course, still valid, but it's of little use: simply because ...

0

The charges could equally well be the other way round and then the direction of the electric field would be reversed. Perhaps the negative charges where placed on the inner conductor and an equal number of positive charges were induced on the inside of the outer conductor? There will be a potential difference across the conductors due to an excess of ...

-1

At r=R, the field strength will be maximum and the field decreases as you go away from it. But that's not the case when you go inside. Any point inside the sphere contains no electric field. The field suffers a discontinuity at the surface. But however the potential is a constant through out inside. It makes sense. The potential is continuous. The potential ...

0

Those formulas, and those graphs are idealizations. In reality, there are no discontinuous fields, just as there are no zero-thickness shells. If you start with an impossible situation, you will calculate impossible and meaningless results. For real situations, situations for which the theory is valid, the field may change quickly, but it does so ...

1

If $\rho$ is a generic physical quantity, e.g. mass density in this case, then spherical symmetry is represented in the form of $\rho = \rho(\lvert \vec r\rvert)$ and not $\rho = \rho(\vec r)$ with $\vec r$ being the position vector of the point at which the quantity $\rho$ is being measured.It is assumed that the center of mass of the distribution coincides ...

0

This is one of those questions which very much relies on the fact that the plane is infinite. So any electric field line which starts off going towards the plane will be counted as part of the electric flux through the plane. Any electric field line which has a z component must reach the infinite z=26 cm plane.

2

A closed surface like a sphere encloses some volume. Anything coming out through the surface (the net outward flow which we call the flux) will be in the expense of what remains inside. If the sphere encloses some charge, then electric field diverging out from the volume containing the charge will be equal to the normal component of the electric field lines ...

0

Not all vector fields have zero flux over any closed surface. Those having zero flux everywhere are a special type called the 'solenoidal'. The electrostatic field doesn't belong to this group. I think you need to study some vector analysis to capture the concepts correctly.

0

Your book just means that the electric field between a pair of parallel plates is uniform, by that it is both constant in direction and value (this is actually only approximately true in practice, there is a small variation). There is a potential difference between the plates because there is an electric field between the plates. Because the field is ...

0

so does it mean that no work is done in moving a point charge towards or away from it? No. A charge under the influence of an electric field has a force exerted on it. So to move the charge work needs to be done. The electric field strength is equal to minus the potential gradient. So if there is an electric field the potential must change. Now ...

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