Capacitor and Dielectric story

Electricity and Magnetism Level pending

A parallel plate capacitor with plate area $A$ and separation $d$ ( $d<<l$ lateral dimensions $)$ is filled with a dielectric of dielectric constant $k=e^{\alpha x}$ where $\alpha$ is a constant $(\alpha>0)$ and plate is at $x=0$ . The capacitor is charged to a potential $V$ .
Calculate the capacitance $(C)$
Answer comes in the form of $C=\frac{\epsilon_{0}A^{\beta}\alpha}{\gamma- \delta e^{-\alpha d}}$
Type your answer as $\beta+\gamma+\delta=?$ Bonus
1) Sketch $C$ vs $d$
2) Obtain the charge on the plates.
3) Obtain the electric field $\vec{E}(x)$

Details and Assumptions
1) Don't forget to solve bonus problem.
2) Remember the $1^{st}$ point of Details and Assumptions .

The problem is not original.

The answer is 3.

2 solutions

A Former Brilliant Member
May 13, 2020

$E(x)=-\dfrac{dV}{dx}=\dfrac{Q}{\epsilon_0\epsilon A}=\dfrac{Q}{\epsilon_0 A}e^{-αx}\implies$

$V=\dfrac{Q}{\epsilon_0 αA}\displaystyle \int_0^d e^{-αx}dx=$

$\dfrac{Q}{\epsilon_0 αA}(1-e^{-αd})\implies$

$Q=\dfrac{\epsilon_0 αA}{1-e^{-αd}}\times V\implies$

$C=\dfrac{Q}{V}=\dfrac{\epsilon_0 αA}{1-e^{-αd}}$ .

So $β=1,\gamma=1,\delta=1$ , and $β+\gamma+\delta=\boxed 3$ .

Sahil P.
May 13, 2020

Capacitors in series add like resistors in parallel, so $C=\frac{\epsilon_{0}A}{\int_{0}^{d} \frac{dx}{e^{\alpha x}}}$ . This yields the desired answer.

Capacitor and Dielectric story

The answer is 3.

2 solutions

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