- #1
diegzumillo
- 177
- 20
I'm studying Landau's Electordynamics of continuous media and, although I like how succinct it is, sometimes it is too succinct! I'm having trouble with a particular passage, so I'll just try to summarize the section up until the part I don't understand.
The topic at hand is electrostatic field for the particular case when the field is two dimensional, i.e. it depends only on, say, x and y. From \nabla . \vec E=0 we can define a vector potencial \vec A such that \vec E=\nabla \times \vec A. We are also imposing that \vec E lies in the xy plane. Now in the text he says that A can be chosen in a way that it is perpendicular to the xy plane, and this is the first thing that I don't understand very well: if the E field does not change with z and has no component in z direction, how can A be anything other than perpendicular to xy plane? I can see no choice.
The main equations of interest here are the Cauchy-Riemann equations, that comes directly from divergence and rotational of E: (I'm also writing the vector potential as a scalar field, since it only has one component)
E_x = - \partial \phi / \partial x = \partial A / \partial y
E_y = - \partial \phi / \partial y = \partial A / \partial x
It can be shown from these relations that the complex quantity w=\phi -iA is an analytic function of the complex argument z=x+iy. This complex function is the core of the technique being introduced. If I understand this so far, that z is not related to the cartesian component.
Here comes my problem. We want to calculate the flux of the electric field through any section of an equipotential line, which is given by \oint E_n dl=-\oint (\partial \phi / \partial n)dl and n is the normal direction. Using one of the above relations we can write this using the potential A:
\oint E_n dl=\oint (\partial A/ \partial l)dl
So far I'm following, but then he writes
\oint E_n dl=\oint (\partial A/ \partial l)dl=A_2-A_1
"where A_1 and A_2 are the values of A at the ends of the section. In particular, since the the flux of the electric field through a closed contour is 4\pi e, where e is the total charge enclosed by the contour per unit length of conductors perpendicular to the plane, it follows that e=(1/4\pi)\Delta A "
He lost me there. The "ends" of a closed contour?
The topic at hand is electrostatic field for the particular case when the field is two dimensional, i.e. it depends only on, say, x and y. From \nabla . \vec E=0 we can define a vector potencial \vec A such that \vec E=\nabla \times \vec A. We are also imposing that \vec E lies in the xy plane. Now in the text he says that A can be chosen in a way that it is perpendicular to the xy plane, and this is the first thing that I don't understand very well: if the E field does not change with z and has no component in z direction, how can A be anything other than perpendicular to xy plane? I can see no choice.
The main equations of interest here are the Cauchy-Riemann equations, that comes directly from divergence and rotational of E: (I'm also writing the vector potential as a scalar field, since it only has one component)
E_x = - \partial \phi / \partial x = \partial A / \partial y
E_y = - \partial \phi / \partial y = \partial A / \partial x
It can be shown from these relations that the complex quantity w=\phi -iA is an analytic function of the complex argument z=x+iy. This complex function is the core of the technique being introduced. If I understand this so far, that z is not related to the cartesian component.
Here comes my problem. We want to calculate the flux of the electric field through any section of an equipotential line, which is given by \oint E_n dl=-\oint (\partial \phi / \partial n)dl and n is the normal direction. Using one of the above relations we can write this using the potential A:
\oint E_n dl=\oint (\partial A/ \partial l)dl
So far I'm following, but then he writes
\oint E_n dl=\oint (\partial A/ \partial l)dl=A_2-A_1
"where A_1 and A_2 are the values of A at the ends of the section. In particular, since the the flux of the electric field through a closed contour is 4\pi e, where e is the total charge enclosed by the contour per unit length of conductors perpendicular to the plane, it follows that e=(1/4\pi)\Delta A "
He lost me there. The "ends" of a closed contour?
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