# Problem on Capacitors.

1 vote
221 views

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

edited May 6, 2018
Welcome to the site Ayush. Please note that we prefer to have the image posted here, instead of a link. You can find instructions in the comments at http://physics.qandaexchange.com/?qa=253/how-to-upload-an-image.

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 (28,896 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.

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)

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