Gauss' law in line integral, Q=##ϵ_0 ∮E.n dl=-ϵ_0 ∮∂ϕ/∂n dl##

In summary, the conversation discusses the use of the Gauss law for surface integrals to calculate total charge on a conductor in a 2D problem. Green's Theorem is used to relate the line integral of the normal components of the electric field to the surface integral, and the outward unit normal is used in the calculations. The individual is seeking clarification on how to calculate the total charge on the conductor using the potential distribution and the Gauss divergence theorem. They propose an equation for the total charge, but are unsure if it is correct.
  • #1
mdn
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I know the Gauss law for surface integral to calculate total charge by integrating the normal components of electric field around whole surface . but in above expression charge is calculated using line integration of normal components of electric field along line. i don't understand this relation. any help please.
 
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  • #2
Is this a 2D problem? Because integrating over a closed curve (ie the "surface" of an area) is the 2D equivalent of integrating over the surface of a volume. By Green's Theorem, [tex]
\begin{split}
\int_{\Omega} \nabla \cdot \mathbf{E}\,dA &= \oint_{\partial\Omega} (-E_y, E_x, 0)\cdot \mathbf{t}\,dl \\
&= \oint_{\partial \Omega} (\mathbf{k} \times \mathbf{E}) \cdot \mathbf{t}\,dl \\
&= \oint_{\partial \Omega} \mathbf{E} \cdot (\mathbf{t} \times \mathbf{k})\,dl \\
&= \oint_{\partial \Omega} \mathbf{E} \cdot \mathbf{n}\,dl
\end{split}[/tex] since [itex]\mathbf{k} = \mathbf{n} \times \mathbf{t}[/itex] where [itex]\mathbf{t}[/itex] is the unit tangent of the curve traversed anticlockwise and [itex]\mathbf{n}[/itex] is the outward unit normal.
 
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  • #3
pasmith said:
Is this a 2D problem? Because integrating over a closed curve (ie the "surface" of an area) is the 2D equivalent of integrating over the surface of a volume. By Green's Theorem, [tex]
\begin{split}
\int_{\Omega} \nabla \cdot \mathbf{E}\,dA &= \oint_{\partial\Omega} (-E_y, E_x, 0)\cdot \mathbf{t}\,dl \\
&= \oint_{\partial \Omega} (\mathbf{k} \times \mathbf{E}) \cdot \mathbf{t}\,dl \\
&= \oint_{\partial \Omega} \mathbf{E} \cdot (\mathbf{t} \times \mathbf{k})\,dl \\
&= \oint_{\partial \Omega} \mathbf{E} \cdot \mathbf{n}\,dl
\end{split}[/tex] since [itex]\mathbf{k} = \mathbf{n} \times \mathbf{t}[/itex] where [itex]\mathbf{t}[/itex] is the unit tangent of the curve traversed anticlockwise and [itex]\mathbf{n}[/itex] is the outward unit norm
yes, this is the 2D problem. i am trying to calculate the total charge on conductor, shown by bold lines in 2D
1664674830832.png

domain. i have calculated potential distribution at each point using Laplace equation in finite element method, now i want to calculate total charge Q on this conductor, from this charge i want to calculate the capacitance. i want to use gauss divergence theorem to calculate charge ##E\bar =-\nabla\phi##, where ##\phi =potential##. here is my real problem. how to calculate total charge on conductor in above case and which normal components of electric field (in terms of scalar potential ##\nabla\phi##) should i integrate to get total charge?

from my side i have written equation like this..
##Q=\oint (E. n) dl##
Q=##\nabla\phi .n##
=##\oint(\partial\phi/\partial x +\partial \phi/\partial y)## dl
=##\oint {\partial \phi/ \partial x} dl +\oint {\partial \phi/ \partial y} dl##
=##\oint {\partial \phi/ \partial x} dy +\oint {\partial \phi/ \partial y} dx##
is this correct?
 

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1. What is Gauss' Law in line integral?

Gauss' Law in line integral is a mathematical equation that relates the electric field at a point to the charge enclosed by a surface surrounding that point. It is a form of Gauss' Law that is used to calculate the electric field in situations where the charge distribution is not symmetric.

2. How is Gauss' Law in line integral expressed mathematically?

Gauss' Law in line integral is expressed as Q=ϵ0 ∮E·n dl=-ϵ0 ∮∂ϕ/∂n dl, where Q is the charge enclosed by the surface, ϵ0 is the permittivity of free space, E is the electric field, n is the unit vector normal to the surface, and ∂ϕ/∂n is the derivative of the electric potential with respect to the normal direction.

3. What is the significance of Gauss' Law in line integral?

Gauss' Law in line integral is a powerful tool in solving complex electrostatic problems. It allows us to calculate the electric field at a point by considering the charge enclosed by a surface surrounding that point, rather than having to consider the individual contributions of each charge in the system. This saves time and simplifies the calculations.

4. What are the conditions for applying Gauss' Law in line integral?

There are two main conditions for applying Gauss' Law in line integral. First, the charge distribution must be continuous, meaning that there are no breaks or gaps in the charge distribution. Second, the electric field must be constant along the line integral, meaning that the electric field does not change in magnitude or direction as you move along the line.

5. How is Gauss' Law in line integral related to Gauss' Law in integral form?

Gauss' Law in line integral is a generalization of Gauss' Law in integral form. While Gauss' Law in integral form is used to calculate the electric field in situations where the charge distribution is symmetric, Gauss' Law in line integral can be used in more complex situations where the charge distribution is not symmetric. In fact, Gauss' Law in line integral reduces to Gauss' Law in integral form when the charge distribution is symmetric.

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