2d Coulomb and Biot-Savart laws

[Tsunami]

I'm trying to find some sort of simple derivation of these laws in 2d, using the integral expressions of the Maxwell equations.

For 2d Coulomb, I found this:

Imagine a rod of infinite length along the z-axis, carrying a charge q which is uniformly divided:

rho (the charge/volume) = lambda*dz*delta(x)*delta(y)

with dq= lambda*dz

So, using the electrical Gauss' law, Er being the value of E in radial direction:

eps0*Er*2*pi*r*dz = lambda*dz

Er= lambda/(eps0*2*pi*r)

Er =-dV/dr

=> V= -lambda/(2*eps0)*ln(r)

So, Coulomb's law would be, with pi(r´) being the electrical charge of the surface evaluated in point r´, en ds´ being the surface that is integrated:

V(r) = -lambda/(2*eps0)* int ( ln(r-r´) , ds´) ?

Does that make any sense? Or is this way off the mark?

And for the Biot-Savart law in the same manner, will the vector potential be something analogous to this thing?

Thank you,

W.

 

[member 553083]

Your derivation of Coulomb's law is correct. For the Biot-Savart law, the expression for the vector potential in two dimensions is: A(r) = (1/(4πε0)) * ∫ ( (r-r') x dl' ) / |r-r'|^3

 

[thepatientmental]

Hello W,

Thank you for your question. It seems like you are on the right track with your derivation of the 2d Coulomb and Biot-Savart laws. However, there are some minor corrections and clarifications that I would like to make.

Firstly, your expression for the electric field (Er) in 2d Coulomb's law should be Er = lambda/(2*eps0*pi*r) instead of Er = lambda/(eps0*2*pi*r). This is because in 2d, the surface area of a circle is 2*pi*r instead of 4*pi*r^2 in 3d.

Secondly, the expression for the potential (V) in 2d Coulomb's law should be V = -lambda/(2*eps0)*ln(r) instead of V = -lambda/(2*eps0)*ln(r-r'). This is because in 2d, the electric potential is only dependent on the distance (r) from the charge, and not on the position (r') of the charge.

In terms of the Biot-Savart law, the vector potential (A) in 2d is given by A = (mu0/(4*pi))*int((J(r')/r')*ds') where J(r') is the current density, r' is the position vector of the current element, and ds' is the surface element integrated over. This is analogous to your expression for the potential in 2d Coulomb's law, but with the addition of the current density term.

I hope this helps clarify your derivation. Good luck with your studies!