# Tag Info

4

The direction of the field is given by the sign of the charge. It can be either outward or inward .

3

Your arguments are not in the right order: First, you prove by symmetry, that the field is radial and depends only on the distance to the center O : $\vec{E}=E(r)\vec{e_r}$. Note that $E(r)$ is an algebraic quantity: you don't need to know its sign. Then, you choose a Gauss surface adapted to this symmetry: here, a sphere centered on O. With this choice of ...

3

Contrary to the comments, this problem can be solved analytically without the use of differential equations. Since this is a "homework-like" problem, but one requiring an unusual technique, I will describe the method in a bit more detail than I would otherwise. I will still leave the details of the calculations to you. Let $x = 0$ denote the ...

2

Any flow of electron signifies current flow. And any flow of current generates magnetic field.

2

Nope. I did this in a first year physics assignment and rightly lost marks for it. The center of charge concept is helpful for finding out how a charged object responds to a uniform field that you already know. To compute it in the first place though, there is no way around the integral. The electric field produced by a charge distribution depends on its ...

2

As you know by Gauss Law, electric field inside a conductor is zero.Thus in what ever orientation the negative charge is induced in the inner surface of the conductor it creates no electric field through the conductor thus it cannot affect the orientation of the positive charge on the outer surface. This is because a charge can only be moved or displaced by ...

2

In the picture you included, there are two distinct elements. One is Gauss’s law proper $$\oint \vec E\cdot d\vec A=\frac{q}{\epsilon_0}\, ,$$ which is always true and works for any closed surface. The other part deals with using Gauss’s law to recover the field, i.e. the step $E=kq /r^2$. This is only possible in some circumstances where the surface and ...

1

Gauss law deals with electric flux and electric flux is related to number of field lines cutting or passing through a surface. And a charge create definite number of field lines, thus if any closed surface is enclosing the charge every field lines emmited or created by the charge will pass through the closed surface. So, whatever the closed surface be, the ...

1

The so-called Faraday rotation, or magneto-optic effect, is an interaction between a magnetic material and linearly polarized EM wave, see https://en.wikipedia.org/wiki/Faraday_effect. The rotation in question is the turning of the linear polarization of the EM wave (light) that is moving within a homogeneous magnetic material and in a parallel direction ...

1

Since the field line indicates the direction of the field, The field at each point must be tangent to the line. In an xy system, dy/dx = $E_y/E_x$. Putting the (+q) at the origin of an xy system you can use Coulomb's law the find an expression for the resultant $E_y$ and $E_x$ as a function of (x) and (y). That gives you (dy) as a function of (dx). I was ...

1

It would be uniformly distributed if the conductor was a sphere but in case of cube there occurs high charge density at corners, thus the charge will be non uniformly distributed. As surface charge density is inversely proportional to area, area at the vertices minimizes to point sized thus charge density is high there. Take this image as an example

1

To find the components of any vector $\bf F$ using unit vectors, you can use the dot product between the vector and each unit vector. So the x-component of $\bf F$ is $\bf F\cdot \hat i$ the y-component is $\bf F\cdot \hat j$ and the z-component is $\bf F\cdot \hat k$ If you have a "direction vector" $\bf u=(x,y,z)$ then its unit vector would be \...

1

Let $\,r\,$ be the radius of a spherical surface with uniform surface charge density $\,\sigma\boldsymbol >0\,$ so with charge $\,q\boldsymbol=4\,\pi\,r^2\sigma\boldsymbol >0$. Then the field inside this spherical surface is zero while the field outside is the same as if the charge $\,q\boldsymbol >0\,$ is concentrated on the center of the sphere so ...

1

One way to approach this is to write out the general solution for Laplace's equation using separation of variables. We assume that the charge distribution on the surface of the sphere $\sigma$ is known (or equivalently the polarization $\vec{P}$ is known, since $\sigma = \hat{r} \cdot \vec{P}$ on the surface), and we solve only for the effects of this ...

Only top voted, non community-wiki answers of a minimum length are eligible