- #1
mitch_1211
- 99
- 1
I have a parallel plate capacitor with a dielectric inside whose relative permittivity varies as εr = βexp([itex]\alpha[/itex]x) [a theoretical calculation, I have not measured this directly]
Eventually I want to calculate the capacitance of the system. First I need to calculate the [itex]\overline{D}[/itex] field, to do this I am using [itex]\oint[/itex]S [itex]\overline{D}[/itex] . d[itex]\overline{s}[/itex] = qfree the result for a parallel plate capacitor is D = q/A with q being qfree again.
Now it can be shown that
[itex]\overline{D}[/itex] = εrε0[itex]\overline{E}[/itex] and from this [itex]\overline{E}[/itex] = D/ε0εr[*]
And this would be fine if the permittivity was constant.
My guess is to assume the dielectric is made up of many infinitesimally small dielectric's each with εr satisfying εr = βexp([itex]\alpha[/itex]x)
I know that an integral will give the sum of many infinitesimal 'pieces' but here is seems as if I have the total (which is the full dielectric) and i need to find each infinitesimal piece of dielectric such that εr = βexp([itex]\alpha[/itex]x) is satisfied for each.
Is that even possible? Will [*] then have an integral on the denominator in place of εr?
I have done a similar calculation for 2 different dielectrics, this seems an extension of this where each dielectric is infinitesimally small.
any help would be much appreciated
Eventually I want to calculate the capacitance of the system. First I need to calculate the [itex]\overline{D}[/itex] field, to do this I am using [itex]\oint[/itex]S [itex]\overline{D}[/itex] . d[itex]\overline{s}[/itex] = qfree the result for a parallel plate capacitor is D = q/A with q being qfree again.
Now it can be shown that
[itex]\overline{D}[/itex] = εrε0[itex]\overline{E}[/itex] and from this [itex]\overline{E}[/itex] = D/ε0εr[*]
And this would be fine if the permittivity was constant.
My guess is to assume the dielectric is made up of many infinitesimally small dielectric's each with εr satisfying εr = βexp([itex]\alpha[/itex]x)
I know that an integral will give the sum of many infinitesimal 'pieces' but here is seems as if I have the total (which is the full dielectric) and i need to find each infinitesimal piece of dielectric such that εr = βexp([itex]\alpha[/itex]x) is satisfied for each.
Is that even possible? Will [*] then have an integral on the denominator in place of εr?
I have done a similar calculation for 2 different dielectrics, this seems an extension of this where each dielectric is infinitesimally small.
any help would be much appreciated