- #1
Apteronotus
- 202
- 0
Hi Everyone,
I'm told that the following formula represents Kirchhoff's current law
[tex]
g_1 E_{1 n}+\varepsilon_1 \frac{\partial E_{1 n}}{\partial t}=g_2 E_{2 n}+\varepsilon_2 \frac{\partial E_{2 n}}{\partial t}
[/tex]
where the first term on each side is Ohm's law and the conductive current
and the second term is the 'displacement current'
To give you the complete picture, we are looking at the boundary of two regions (1 and 2) having different dielectric properties.
[tex]E_{i n}[/tex] is the normal component of the electric field in region i
[tex]\varepsilon_i[/tex] the dielectric constant there, and
[tex]g_i[/tex] the conductivity
Could someone please shed some light on this. What exactly is the second term[tex]\varepsilon_i\frac{\partial E_{i n}}{\partial t}[/tex]? What's its physical significance? Is it the buildup of charge on the boundary? If so why do we consider the normal component [tex]E_{i n}[/tex]?
Please make me understand :(
Thanks
I'm told that the following formula represents Kirchhoff's current law
[tex]
g_1 E_{1 n}+\varepsilon_1 \frac{\partial E_{1 n}}{\partial t}=g_2 E_{2 n}+\varepsilon_2 \frac{\partial E_{2 n}}{\partial t}
[/tex]
where the first term on each side is Ohm's law and the conductive current
and the second term is the 'displacement current'
To give you the complete picture, we are looking at the boundary of two regions (1 and 2) having different dielectric properties.
[tex]E_{i n}[/tex] is the normal component of the electric field in region i
[tex]\varepsilon_i[/tex] the dielectric constant there, and
[tex]g_i[/tex] the conductivity
Could someone please shed some light on this. What exactly is the second term[tex]\varepsilon_i\frac{\partial E_{i n}}{\partial t}[/tex]? What's its physical significance? Is it the buildup of charge on the boundary? If so why do we consider the normal component [tex]E_{i n}[/tex]?
Please make me understand :(
Thanks