What is the integral form for the Gauss Law for Electric Fields?
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[Physics] Gauss Law for Electric Fields
electric-fieldselectrostaticsgauss-law
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Let me state Gauss law very briefly:
The surface integral of electric field over a closed surface is equal to the charge contained within the surface over $\epsilon_0$. $$ \int_{\Delta s}{\mathbf{E\cdot ds}} = \frac{q_{en}}{\epsilon_0} $$ $$or$$ The divergence of Electric field at a point of space is equal to the charge density at that point over $\epsilon_0$. $$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$
Unlike Coulomb's law, which directly gives an expression for the electric field, Gauss law gives a differential equation of vectors to solve the electric field from. In short, electric field is difficult to calculate from Gauss law in most circumstances.
But we still can find the electric field if we somehow solve the integral. This is practical when there exists symmetry in the situation where we want to calculate electric field.
So,
for a given charge distribution,I can draw a sphere , a cylinder , a cube and similarly any Gaussian surface enclosing the charge completely...
If you take an arbitrary gaussian surface, chances are that it is not very symmetrical and you are not able to calculate the integral on the left side of the equation. Once you know the electric field, and you now calculate the integral, you'll see that Gauss law holds. So, although you have an infinite number of choices of surfaces to verify Gauss law, you don't have as many to find the electric field from.
but all of them would give different values of electric field right at the same given point.
Let me stress again that this simply isn't true. If you are able to find the electric field from one of the surfaces, and you plug that value value of electric field in Gauss law for any arbitrary surface, then you will always find that Gauss law holds good.
Take home message: You may take a closed surface of any shape as a gaussian surface and apply Gauss law, and there will be no inconsistency.
Hope it makes sense. :)
The statements.
Consider an electric field along the x axis that varies as follows E(x,y,z) = K/(x^2)
and
assume the field vector to be along the positive direction of the X axis, the field lines are parallel and equally spaced, assumed to come from a very large distance
cannot be simultaneously true.
In effect you have proved that with your evaluation of the electric flux through the opposite faces of a cube and showing it to be different.
If the electric field lines are parallel then the electric field is uniform and hence cannot depend on position $x$.
Best Answer
In CGS, we have that $$\nabla\cdot\mathbf{E}=4\pi\rho_{T}$$ Integrating in a volume $V$, we have: $$\iiint_{V}\nabla\cdot\mathbf{E}dV=4\pi\iiint_{V}\rho_{T}dV=4\pi Q_{T}$$ And with the Gauss-Ostrogradsky theorem:
$$\iiint_{V}\nabla\cdot\mathbf{E}dV=\iint_{\partial V}\mathbf{E}\cdot\mathbf{\hat{n}}dS$$ And finally, we get $$\mathbf{\iint_{\partial V}E}\cdot\mathbf{\hat{n}}dS=4\pi Q_{T}$$ This is the correct forms of the Gauss law in cgs. In MKS we have: $$\nabla\cdot\mathbf{E}=\rho/\epsilon_{0}$$ And with the same procedure, we arrive to the result $$\mathbf{\iint_{\partial V}E}\cdot\mathbf{\hat{n}}dS=Q_{T}/\epsilon_{0}$$ Where $Q_{T}$ is the total charge inside the volume $V$.