# Problem on Capacitors.

1 vote
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For following I want to find $E(x)$: I got $Q(x)=\dfrac{\alpha V A\in_{o}}{e^{-\alpha x}(1-e^{-\alpha d)}}\implies E(x)=\dfrac{\alpha Ve^{\alpha x}}{1-e^{-\alpha d}}$.

But solution say that, $E(x)=\dfrac{\alpha Ve^{-\alpha x}}{1-e^{-\alpha d}}$.

Please help me with this, is that solution a misprint or mine result is wrong?

asked May 5, 2018
edited May 6, 2018
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1 vote

Placing a dielectric of constant $k$ between the plates of the capacitor reduces the electric field between the plates and increases capacitance, both by factor $k$.

The electric field in the capacitor should be inversely proportional to dielectric constant : $E(x) \propto 1/k(x)$. In your solution the electric field is directly proportional to dielectric constant ($E(x) \propto k(x)$) so it is incorrect.

answered May 5, 2018 by (27,556 points)
selected Jul 2, 2018 by n3
$E=\dfrac{Q}{A\in_{o}}$ (between parallel plates) gives me what I wrote above, if $Q(x)=\dfrac{\alpha V A\in_{o}}{e^{-\alpha x}(1-e^{-\alpha d)}}$.
What is $Q(x)$? The capacitor with variable permittivity $k(x)$ can be modelled as an infinite number of capacitors each with fixed permittivity **connected in series**. The charge on each of these capacitors is the same, it does not depend on $x$.
$Q(x)$ is the charge on capacitor kept at $x$ from an origin.
Sorry I don't understand what you mean by your definition of $Q(x)$. The charge on the single capacitor with variable permittivity has the same magnitude on both plates, $+Q$ on one plate at $x=0$ and $-Q$ at the other plate at $x=L$. When modelled as an infinite series of capacitors the charge on each capacitor is the same.

Would you like to post your solution, either in your question or as an answer, explaining the steps in your calculation?
1 vote

net capacitance of the system is series combination of "elemental capacitors" each of thickness $dx$

$\dfrac{1}{C_{net}}=\dfrac{1}{A\epsilon_{0}}\displaystyle \int_{0}^{d}\dfrac{dx}{e^{ax}}\implies C_{net}=\dfrac{aA\epsilon_{0}}{1-e^{-ad}}$

$V=\dfrac{Q}{C_{net}}=Q\left[\dfrac{1-e^{-ad}}{aA\epsilon\_{0}}\right]\tag{1}$

by inserting inserting dielectric D remains same

we have ,

$\vec {D}=electric \ displacement=\epsilon_{0}K E \ { {\hat x}}$

again by definition of D i.e, it equal to the amount of free charge on one plate divided by the area of the plate $\left(\dfrac{Q}{A}\right)$

$D=\left(\dfrac{Q}{A}\right)=\epsilon_{0}K E \ { }\implies E=E(x)=\dfrac{Q}{A\epsilon_{0}{K}}$

in above, put$K=e^{ax}$

$E(x)=\dfrac{Q}{A\epsilon_{0}e^{ax}}$

if you calculate $-\displaystyle \int _{0}^{d} E(x) dx$ using above expression of E(x) you will get same voltage as above eq.(1)

i think answer in your book is wrong

and also, charge $Q(x)$ will not vary as function of $x$because both $C_{net}$ and $V$ are fixed constant quantities

answered Sep 7, 2018 by (120 points)
edited Sep 7, 2018 by Einstein