- #1
godtripp
- 54
- 0
This isn't a homework problem, just something to explore my understanding so correct me if any of my assumptions are wrong.
I have given myself the situation where I have a capacitor of area "A" and separation "d"
Between them is a non-uniform but continuous dielectric, so that the dielectric on the right plate is significantly more dense than that of the left.
I have also given the dielectric the function: [tex]\epsilon = \epsilon_{0}+\alpha x[/tex]
where x is the distance from the left plate.
Now I can find my answer by taking an infinite sum of infinitesimal capacitors in series.
I'm just confused now on how to properly build the formula. Should I take the derivative of
my capacitance formula with respect to d, plug in for epsilon as a function of distance and integrate? Or how?
I've listed a few of my ideas on formulas below. If anyone can point out the right one and why.
[tex] C = \frac{\epsilon_{ave} A}{d} [/tex]
Where
[tex]\epsilon_{ave}=\frac{1}{d}\int \epsilon_{0}+\alpha x dx[/tex]
[tex] C_{eq} = \int \frac{-(\epsilon_{0}+\alpha x)A dx}{x^2} [/tex]
[tex] \frac{1}{C_{eq}}= \int \frac{dx}{(\epsilon_{0}+\alpha x)A} [/tex]
[tex] \frac{1}{C_{eq}}= \int \frac{x dx}{(\epsilon_{0}+\alpha x)A} [/tex]
I have given myself the situation where I have a capacitor of area "A" and separation "d"
Between them is a non-uniform but continuous dielectric, so that the dielectric on the right plate is significantly more dense than that of the left.
I have also given the dielectric the function: [tex]\epsilon = \epsilon_{0}+\alpha x[/tex]
where x is the distance from the left plate.
Now I can find my answer by taking an infinite sum of infinitesimal capacitors in series.
I'm just confused now on how to properly build the formula. Should I take the derivative of
my capacitance formula with respect to d, plug in for epsilon as a function of distance and integrate? Or how?
I've listed a few of my ideas on formulas below. If anyone can point out the right one and why.
[tex] C = \frac{\epsilon_{ave} A}{d} [/tex]
Where
[tex]\epsilon_{ave}=\frac{1}{d}\int \epsilon_{0}+\alpha x dx[/tex]
[tex] C_{eq} = \int \frac{-(\epsilon_{0}+\alpha x)A dx}{x^2} [/tex]
[tex] \frac{1}{C_{eq}}= \int \frac{dx}{(\epsilon_{0}+\alpha x)A} [/tex]
[tex] \frac{1}{C_{eq}}= \int \frac{x dx}{(\epsilon_{0}+\alpha x)A} [/tex]