# Force between two capacitor plates

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A parallel plate capacitor with air as a dielectric is arranged horizontally. The lower plate is fixed and the other connected with a vertical spring. The area of each plate is A. In the steady position, the distance between the plates is d$_0$ . When the capacitor is connected with an electric source with the voltage V, a new equilibrium appears, with the distance between the plates as d$_1$ . Mass of the upper plates is m.
(i) Find the spring constant K.
(ii) What is the maximum voltage for a given K in which an equilibrium is possible ?
(iii) What is the angular frequency of the oscillating system around the equilibrium value d1 .(take amplitude of oscillation << d1 )

I tried as

Electroc field due to one plate is $(Q)/(2A\epsilon)$

Hence force between two plates is $(Q^2)/(2A\epsilon)$

asked Feb 6, 2017
Try balancing the forces on the upper plate. Note that $Q=CV$ and that $C$ and $V$ both change.
But in that i don't know charge
Do you have a worked solution which you are trying to understand? If so, please can you upload it or provide a link.
Sorry I only have the answer  http://i.imgur.com/6n5gAHc.jpg

(i) The extension of the spring is $(d_0-d_1)$ so the additional upward force exerted by the spring at equilibrium (in excess of the force $mg$ required to support the uncharged plate) is $K(d_0-d_1)$.

At equilibrium the electric potential between the plates is $V$ and the separation is $d_1$. The electric field is uniformly $E=V/d_1$. The charge on each plate is $Q=CV=\epsilon A V/d_1$. Each plate contributes $\frac12 E$ to the electric field between the plates. Each plate only experiences the electric field due to the other plate. So the force of attraction pulling the top plate down is $\frac12 QE=\epsilon AV^2/2d_1^2$.

Equating these two forces gives
$K=\epsilon AV^2/2(d_0-d_1)d_1^2$.

(ii) Rearrange the force-balance equation to get
$V^2=B(d_0d_1^2-d_1^3)$
where $B$ is a constant. Then differentiate to find the maximum value of $V$ as a function of $d_1$.

(iii) Suppose $d_1$ increases by a small amount $x$. Then the upward force on the upper plate from the spring becomes $K(d_0-(d_1+x))$ and the downward force from the fixed lower plate becomes $\epsilon AV^2/2(d_1+x)^2 \approx (\epsilon AV^2/2d_1^2)(1-2\frac{x}{d_1})$ using the binomial expansion and assuming that $x \ll d_1$. The equation of motion of the upper plate is then
$m\ddot x = K(d_0-d_1-x)-\frac{\epsilon AV^2}{2d_1^2}(1-2\frac{x}{d_1})$
$m\ddot x +(K-\frac{\epsilon AV^2}{d_1^3})x=K(d_0-d_1)-\frac{\epsilon AV^2}{2d_1^2}=0$
because the forces balance at equilibrium separation $d_1$. This equation is of the form $\ddot x + \omega^2x = 0$ so the motion is (approx) SHM with angular frequency $\omega$, where
$\omega^2 =\frac{K}{m}-\frac{\epsilon AV^2}{md_1^3}$.

answered Feb 6, 2017 by (28,746 points)
edited Jul 19, 2018
Good point, I got it wrong. Velut luna's answer is the best. I will correct my answer.