3

Let $\vec{r}=x\hat{x}+y\hat{y}=r\hat{r}$ and $\vec{\tilde{r}}=\tilde{x}\hat{x}+\tilde{y}\hat{y}$. Then the integrand becomes $$\frac{\vec{r}-\vec{\tilde{r}}+z\hat{z}}{(|\vec{r}-\vec{\tilde{r}}|^2+z^2)^{3/2}}$$ Since the square of charge is far away from the origin, when the point $(x,y,z)$ is very far from the square of charge, it will also be very far ...


3

You are right in saying that the electric field near a point charge is infinite, but that is not the case for continuous volume charge distributions. The best example is a charged solid sphere. Although the solid sphere has charge, still you can define electric field inside the sphere at any point and this electric field is finite, not infinite. And thus if ...


3

Gauss' law relates the divergence of the electric field to the charge density. If you are considering the exact point at which a charged particle is "located" you are considering a point of zero volume, within which the charge density is infinite. If you take a charged particle at a point $p$ you find that $$\rho(p)=\lim_{V\rightarrow 0}\frac{Q}{V}=\infty.$$...


2

The fibres are made from a high resistivity synthetic polymers called electrets which can maintain a permanent dipole moment both on the surface and in the bulk. In many ways they are the electrostatic equivalent of permanent magnets. They are charged with excess charge on the surface and by producing permanent bulk dipoles in the manufacturing process ...


1

By taking the line of charge to be infinite, we've introduced the fact that translation along the y axis is invariant; if we measure the field around the wire at some point, then move along the wire, we expect to see the same field as our system still looks the same, i.e. we have no way of measuring our 'y-coordinate' along the wire. Now by symmetry, if we ...


1

No, because the surface charge density is not necessarily uniform. In 2D, the amount of charge on a small line element $ds$ on the boundary is $\lambda ds$, where $\lambda$ is charge per length on the boundary. The force on it is then $\frac{1}{2}\lambda \vec{E} ds$. (The factor of $\frac{1}{2}$ arises from the fact that the net electric field "felt" by ...


1

I'm pretty sure that this problem is solvable by using image charges. However, you have to make sure that you incorporate the images charge(s) of plate 1 into the calculation for the image charge(s) of plate 2. Hence, one obtains an infinite series of image charges. I'd like you to do the calculation and post your answer here.


1

The increment of work done by a force is $dW = {\bf F} \cdot d\bf x$. The force on a charge is ${\bf F} = q {\bf E}$, where the electric field is ${\bf E} = -\nabla \phi$. That means that the increment of work done on a charge by the electric force is $dW = -q\nabla \phi \cdot d{\bf x} = -q d\phi$. So, by moving from a lower potential to a higher ...


1

Say, you've got a negative charge and a plate near it. Now suppose that the plate has a positive charge. So, you can say that the "location" of the positively charged plate is at a higher potential than the "location" of the negative charge. Also, it is obvious that the positive plate will attract the electron towards it, and the electron would move from ...


1

Yes, your surface integral would change depending on the shape of the surface you're talking about, if you take a highly complex surface you're going to have to solve a highly complex surface integral. Gauss' law applies in all cases though. If you have a closed surface, the flux of the electric field through it is always proportional to the charge contained ...


1

why does not the dielectric field cancel out the capacitor's field? The polarization of the dielectric in the capacitor does reduce the effective electric field of the capacitor, but doesn't completely cancel it out. The reason is the molecules of the dielectric material are not perfectly polarized by the capacitor's electric field. See the diagrams below ...


1

The charges in the dielectric to rearrange and lessen the field inside of the capacitor, the field just isn't completely canceled. This is because, as you said, we are dealing with bound charges. If you compare a vacuum filled capacitor with charge $\pm Q$ on its plates to a capacitor with the same charge filled with a linear dielectric of dielectric ...


1

Based on the removed comments I found that the error I made was due to me taking a volume integral rather then the surface integral. Taking the surface integral (2) becomes $$Q=\epsilon_0 E(r=\frac{1}{\alpha} )\frac{1}{\alpha^2}\int_0^\pi \int_0^{2\pi} sin(\theta )d\theta d\phi\tag{2} = 4 \pi \epsilon_0 c(1-e^{-1})$$ The additional factor of $4 \pi \...


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