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0

I think it's worth approaching your question from a bit of a different point of view. We know that the electric field is a vector field. In general, the divergence of a vector field is a measure of field sources/sinks. This is just an interpretation of a mathematical property, and applies to any vector field. We'll use the symbol rho to signify the field ...


0

Other answers have already said why that should be zero physically. But maybe you are interested in mathematical proof. You can easily proof that by using Coulomb's law $\vec{E}=\frac{1}{4\pi\epsilon_0} \frac{1}{r^2}$ and solid angle. Take a point charge from the charges outside the surface. Find out the flux due to it. In the following figure see how the ...


0

You write, "In the case of the magnetic field we are yet to observe its source or sink." If you mean "we are yet to observe a source or sink", you're correct. However, consider the magnetic vector field (ignoring units/speaking qualitatively): $$\vec{B}=(0, \frac{z}{(1+r^2)^2},\frac{y}{(1 + r^2)^2})$$ This is a valid field because it's the curl of the ...


10

This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law \begin{equation} \nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0} \end{equation} If we integrate this over some volume $V$ and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} ...


6

I) Right, the differential form of Gauss's law $$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0} $$ uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges $$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$ Note in particular, that it is technically wrong to claim (as ...


1

Maxwell's equations state $$ \nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$$ $$ \nabla \cdot \vec B = 0 $$ If we accept Maxwell's equations as true, there is no source/sink of the magnetic field, since the divergence of the magnetic field is zero no matter what. Yet, no matter how you feel about the Dirac delta, where there is charge, there is non-zero ...


2

The argument is perfectly sound. Hand waving is a diminutive term for this, since this really is the essence of Gauss' Law. Here is a succinct physical answer, without using maths, which you can use to make sense of the situation. Field lines due to charges located outside will cross the surface first while ''going in'' and then, while ''going out'', and ...


2

Your "conservation of field lines" argument is solid, otherwise you have basically just used Gauss's law in your work above. By the divergence theorem and Maxwell's equations, $$\int \vec{E}\cdot \vec{dA}=\int \vec{\nabla} \cdot \vec{E} dV=\int \rho /\epsilon_0 ~dV=\rho_{enc} / \epsilon_0$$ So the (flux of) electric field only depends on the enclosed charge. ...


0

This is an extremely common mistake in introductory EM - from students who actually spend time thinking about the problem, anyway ;-) Use Gauss's law in both cases: In the case of infinite plates, you do not have the result you give first. A Gaussian cylinder has two disks on either side of the plate, so $$E_1(2A)=\frac{\sigma A}{\epsilon_0}\rightarrow ...


1

Pick a point above the plane. From a point in the plane directly under the point above, draw a circle of some radius. Consider the contribution of the charge elements along the circle to the electric field at the point above the plane. Since the charge density is uniform, the horizontal components of the electric field from charge elements on opposite ...


1

An answer connected to Gauss law (I hope everything is correct, since it's long ago for me ... so no warranty): An infinite plane of uniform charge for example in the z-plane has the charge distribution: $\rho=q\,\delta(z)$ Thus, the electrostatic potential should be $\Phi=\frac{q\,|z|}{2\pi}$. Hence, the electric vectorfield is: ...


8

The answer by @NowIGetToLearnWhatAHeadIs is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation. Start with your charge distribution and a "guess" for the direction of the electric field. As you can see, I made the guess have a component upward. We'll see shortly why ...


0

With respect to your test charge, there will always be an equal number of charges on your plane in all directions because the plane is infinite. So for every charge "in front" of your test charge there will be a charge "behind" your test charge. And for every charge to the left of your test charge will be a charge to the right. What this means is that no ...


8

You have to realize that the system is invariant under rotations about the normal to the plane. Then then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to ...


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In addition to Qmechanic answer, you might be interested to look this page too http://physicspages.com/2011/11/14/dirac-delta-function-in-three-dimensions/.


2

Let us ask the question the other way round. Given a conductor closed shape and a zero field inside, what are the possible surface charge distributions and when is one of them constant ? The potential is solution of the Laplace equation and can written as $$V(r,\theta,\phi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \left(A_{\ell m}r^\ell+B_{\ell ...


7

It doesn't hold for arbitrary shapes. The reason it works for spheres is that when you have a spherical charge distribution and a concentric spherical Gaussian surface, the whole system is invariant under rotations around the center of the spheres. If the electric field were different at different points on the Gaussian sphere, you could rotate the whole ...


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For a convex surface, take any arbitrary point inside the closed surface. Now consider a cone with infinitesimal angle from that point so that it cuts the surface on both side.now you calculate the electric field for these infinitesimal surfaces say $\mathrm{d}s_1$ and $\mathrm{d}s_2$. You will find electric field as zero. as these two surfaces make same ...



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