# Capacitor with dielectric as spring

#### Muthumanimaran

1. The problem statement, all variables and given/known data
Two parallel plates of metal sandwich a dielectric pad of thickness d, forming an ideal
capacitor of capacitance C. The dielectric pad is elastic, having a spring constant k. If an
ideal battery of voltage V across its terminals is connected to the two plates of this
capacitor, the fractional change $\frac{\delta{d}}{d}$ in the gap between the plates is

2. Relevant equations
$$C=\frac{\epsilon{A}}{d}$$
$$U=\frac{1}{2}CV^2$$
$$U=\frac{1}{2}k\delta{d}^2$$
3. The attempt at a solution

The Capacitance of the capacitor is $U=\frac{1}{2}CV^2$, when the dielectric is elastic the workdone in compressing/elongating the material of spring constant 'k' is $U=\frac{1}{2}k\delta{d}^2$, the fractional change in the energy of the capacitor is $\frac{1}{2}CV^2-\frac{1}{2}k\delta{d}^2$, I don't know what I have done is correct or not and I don't know how proceed from here as well. Please tell me whether my approach is not correct or give me a hint to solve the problem.

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#### scottdave

Homework Helper
Gold Member
I am pretty sure that you cannot equate change in electric field energy to change in spring energy. Perhaps the way to approach is to look at the force exerted on the plates, by the electric field.

#### rude man

Homework Helper
Gold Member
Tricky problem since you must consider the fact that the battery gives or absorbs (you decide) energy also. So the energy balance equation is
old field energy + contributed battery energy + work done by force = new field energy + spring energy.
Hint: battery energy added = V ΔQ.

PS I couldn't decipher your latex in your section "The attempt at a solution".

Last edited:

#### ftRohan

Due to presence of voltage V, the force between the plates of the capacitor will be equal to the restoring force of the dielectric pad of spring constant k. The dielectric displaces by a fractional distance ∆d(∆ represents small change)
Therefore the equation is,
Q^2/2A€ = k∆d (1)

Also, Q=CV (2)

Therefore, eqn(1) implies-
(CV)^2 = 2k∆d×A€ (3)
Also, C=A€/d (4)

Therefore, eqn(3) implies-
(CV)^2 = 2k∆d ×Cd
Dividing both sides by d^2, we get
(1/2 CV^2)/d^2 = k∆d/d

=> ∆d/d = (1/2CV^2)/kd^2.

€ stands for epsilon. Sorry couldn't find the right notation for epsilon.
I hope this helps.

#### gneill

Mentor
Hello ftRohan,

Welcome to Pysics Forums!

Please note for the future that it is against Forum rules to post complete or near-complete solutions to homework problems before the Original Poster has arrived at a correct solution by their own efforts. Helpers can only provide guidance via hints or pointing out errors in the OP's attempts.

"Capacitor with dielectric as spring"

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