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## New answers tagged gauss-law

0

Integrate to get $Q(r)=\int_0^rdq$ next apply Gauss law to get E(r). Use espherical coordenates to integrate dq

2

It is really a matter of combinations. Potential energy is a feature of a system, so between two particles there is one potential energy. The summation however, will cadd the potential energy between two particles twice (e.g., $q_1\phi(\mathbf{r}_2)$ and $q_2\phi(\mathbf{r}_1)$). Hence, the one half term has to be introduced so that the potential energy of ...

3

Your expression for the electric field in Cartesian coordinates is incorrect. The field is not radial, it points away from the line of charge. If we put the line of charge along the z-axis, then the E-field is $$\vec{E} = \frac{x\vec{i} + y\vec{j}}{x^2 + y^2}$$ with some multiplicative constant involving the charge per unit length and $\epsilon_0$. The ...

1

We need to be using cylindrical coordinates instead of the spherical coordinates you have implicitly used. The field you have takes the form: $$\vec E=\frac{\lambda}{2\pi r \epsilon_0} \hat e_r$$ Where we have used cylindrical coordinates. The div in cylindrical coordinates is given by: $$\nabla \cdot (A_r \hat e _r+A_\theta \hat e_\theta +A_z \hat e_z) ... 0 Appending to the answer given above, since I cannot post comments. Integrating in the onion peel manner, charge enclosed in the gaussian surface of a 'thin' shell taking care of limits (inner radius and beyond to whatever distance so desired) should yield you the answer to your question. Note the exponent of the distance parameter in your 'distance - ... 0 To answer the first question, consider that if the electric field acts in one dimension, then \mathbf{E} = -\frac{d\mathbf{V}}{dx} Hence, in a uniform field the potential changes linearly across the direction parallel to the field. Like this image, where the potential increases by a constant amount at each equipotential as we go left (because field ... 0 Gauss's Law states$$\int\limits_{\partial V}\vec{E}\cdot\textrm{d}\vec{S}=\frac{1}{\epsilon_0}\int\limits_V\rho\textrm{d}V$$Then by if the magnitude of the electric field is a constant along the surface \partial V, and theta is the angle between the unit normal to the surface and the electric field, you may calculate$$\int\limits_{\partial ...

1

The basic concept is to see the charge distribution among them. From these example I hope your doubt will be quite clear First left us take one non-conducting sheet then, $$E ds + E ds + 0 = \frac{1}{E_o}[\sigma.ds]$$ For curved surface it will be zero and let us take sigma distribution on sheet. See Figure 1. Now for conducting sheet (Case 1) Here ...

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"So if I construct a gaussian surface such that it extends out from both the sides ,then I should get the same answer as that for a charged sheet! Why do we not do this in this case ?" if you do that you have to realize that the other side have same amount of charge as the first, so the filed is doubled As for your last question if you consider a ...

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An easy way to explain it is a light source emitting light in all directions. The light is the area of a sphere, and volume has nothing to do with it.

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Flux is proportional to the area of the sphere not the volume of the sphere. It is evident from definition of the flux $\Phi_\mathbf{B}$ of some quantity $\mathbf{B}$ , which is defined in the following way, $$\Phi_\mathbf{B}= \iint\mathbf{B} \cdot \mathrm d \mathbf{A}$$ Therefore the flux is proportional to the area of the sphere and hence the $1/r^2$ ...

1

If the electric charge density of a region of space is negative, that would mean that there are more negative charges than positive charges in that region. When people use the word "density" casually, they usually mean mass density (or sometimes number density). Mass (as far as we know) can only be positive, and the number of particles can only be ...

0

It tells us that the charge is negative. Surface charge density is negative when the surface is covered by -ve charge. Same goes for volume charge density. Like any density, charge density can depend on position, but charge and thus charge density can be negative. Wikipedia

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The point is: what do you call a "conductor"? If you are talking about a medium which follows Ohm's law $\vec{j} = \sigma \vec{E}$, then $\vec{E}$ has no reason to equal zero (well, in steady state, $\vec{j}=\vec{0}$ so...). But if you are dealing with a perfect conductor, in which $\sigma \rightarrow +\infty$, then $\vec{E} = \vec{0}$: $$\begin{cases} ... 0 Then there would be electric field inside that shell. Well if it is still intact. Because you need an immense positive charge to do that to a typical shell, more than 100000 Coulombs (assuming you have 1 mole of conductor with about 50 electrons per atom). To bring that much charge together requires immense force and energy. In fact potential energy of 10 cm ... 1 If there were to exist an electric field in the conductor, there would also have to be a current. A radial field drives a radial current, and it is this very current that will change the charge that is on either surface, which then in turn creates its own electric field. The steady-state is when the TOTAL field is equal to zero, when there are no longer any ... 0 The significance of Gauss law is to calculate electric field at a point. This is easily done by making a gaussian surface to pass through the point at which field is required and simultaneously enclosing the charge due to which the field is required inside the gaussian surface. If the gaussian surface is made to pass through a discrete charge itself then the ... 1 Since you know that the field inside a conductor is zero, you can apply Gauss' Law for flux to say that any spherical surface lying inside the conductor cannot have any flux through it, so that the enclosed charge is zero. Therefore, you need -2.3mC of charge from the conductor on the inner surface to make sure that the enclosed charge is zero. For the ... 0 Just because a surface has zero electric flux through it, it does not mean that the electric field will be zero throughout it. What this means is that the field lines that go into the surface must come out of it at some other point. Thus, if \mathbf E\cdot \mathrm d\mathbf S is nonzero at some point in the surface, there must be some other point in the ... 0 Gauss's Law states that,$$\int_S E.dA=\frac{q}{\epsilon_o} The $q$, here, is the charge enclosed inside the Gaussian surface. You are right when you say that the net flux through $S_2$ is zero, but that only means that the net flux due to enclosed charges in $S_2$ is zero. You cannot make implications about the Electric Field due to an external charge ...

1

Gauss law is useful only in the cases of high symmetry systems like sphere, infinitely long (or very long and thin) cylinder, or infinite plane. You can't even apply it to curved sides of "real", short cylinder. You need to show somehow that gravitational field is the same everywhere on you Gauss surface, so that integral in the Gauss law turns out to be a ...

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