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Recently I came across this problem :

There are two identical parallel plates of length $L$ and breadth $B$ on the XZ plane . One plate passes through $Y = 0$ and the other passes through $Y = d$. Between them is a dielectric medium whose Dielectric constant changes as $\kappa = \kappa_0(3 + \frac{y}{L})$. Calculate the Capacity of the parallel plates.

two parallel plates in the XZ plane with a dielectric material between them

The way to solve this problem is by considering a small portion of thickness $dy$ at a distance $y$ from the XZ plane.

Then we use the formula of $C = \frac{\kappa\epsilon_0 A}{d}$ to get the capacitance of the small portion. We denote this capacitance as $dC$

$$\begin{align} \Rightarrow dC &= \frac{\kappa \epsilon_0 LB}{dy} \\ \Rightarrow dC &= \frac{\kappa_0(3 + \frac{y}{L}) \epsilon_0 LB }{dy} \end{align}$$

We then take reciprocal on both sides and integrate it from $Y = 0$ to $Y = d$. The logic behind this is that since the layers of the dielectric are in series therefore we find the equivalent capacitance by integrating the reciprocals of capacitance of individual layers.

$$\Rightarrow \int\frac{1}{dC} = \int\frac{dy}{\kappa_0(3 + \frac{y}{L}) \epsilon_0 LB}\tag{i}$$

Solving the integral we get some value for $1/C$.

But the question is that this step might be logically correct but mathematically wrong because there is no meaning in $\int\frac{1}{dC}$ (or does it?). Can we really integrate undefined values by simply assigning some logical meaning behind it?

David Z
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Jdeep
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  • I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. – David Z Jul 19 '20 at 08:36
  • You could always write things in terms of elastance and then this issue never arises. – Michael Seifert Feb 10 '21 at 14:23

3 Answers3

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As a mathematical notation it is meaningless. But the overall derivation is correct, as you really pass to the continuous limit when switching from summing small elements $1/\Delta C$ to integration over $dy$. There are actually fine differences between an increment, a differential and a derivative, which textbooks often omit in this type of calculations.

A practical tip: write always $\Delta X$ for a small element and a summation sign $\Sigma$, and pass to $dx$ and integral $\int$ only at the last step of the derivation, when you actually take the number of the elements to infinity and pass to integration.

Roger V.
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Your basic mistake is that you didn't do the differential correctly:

$dC= -\frac{3\kappa_0\epsilon_0 A}{y^2}dy$

You need to be careful when you integrate: because both sides of the integral diverge when $y$ goes to zero. The limits of integration for $C$ are $(3\kappa_0\epsilon_0 A/\epsilon, C_d)$, and those of $y$ are $(\epsilon, d)$. Once you do the integrals on both sides you can take the limit $\epsilon \rightarrow 0$ and get a finite result.

  • How did you get this result ? – Jdeep Jul 19 '20 at 15:19
  • that is how you differentiate, take the derivative and multiply by the differential of the independent variable https://en.wikipedia.org/wiki/Differential_(infinitesimal) –  Jul 19 '20 at 15:21
  • But your integral gives a divergent value for C – Jdeep Jul 20 '20 at 17:21
  • You can check the integral here https://www.integral-calculator.com/#expr=%283%2By%2Fl%29%20%2Fy%5E2&intvar=y&lbound=0&ubound=d – Jdeep Jul 20 '20 at 17:27
  • I updated the answer, let me know if you can follow it –  Jul 20 '20 at 18:04
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You can not write such an expression. Just consider the case were $dC$ and $dy$ are infinitesimal small. The left hand side of your expression is small, while the right hand side blows up. Therefore, this expression can not be true.

The correct way of doing such a thing is to write the capacitance as a function of the distance,$C = C(y)$, and take the derivative $$ \frac{dC}{dy} = ... $$ Finally, you multiply both sides with $dy$. Although the last step is mathematically questionable, you end up with the correct expression for $dC$. Finally consider the case of "many" capacitance in series.

Semoi
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    Downvoting because of phrasing (v1). "It's absolutely wrong" is too strong a sentence, considering that the OP's approach was correct modulo the continuous limit. On the contrary @Vadim's answer acknowledges the OP's derivation and points out the correct formulation. – ɪdɪət strəʊlə Jul 19 '20 at 08:52
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    @idiotstroszek: I often wonder why people downvote. Your comment helps me to improve the answer. I'll go right at. Thank's! – Semoi Jul 19 '20 at 08:59
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    @Semoi Indeed, I ask myself the same question whenever I see an answer of mine downvoted for no reason (apparent to me). – Roger V. Jul 19 '20 at 09:06
  • @Semoi it is now better (v2), but I still feel that "You can not write such an expression" is a bit strong. For one, they just did, but also, within an appropriate context (e.g. differential forms and 1/ meaning inverse in some sense), such an expression could be reasonable. In the current context, indeed it is meaningless, but that doesn't mean they can't write it :) – ɪdɪət strəʊlə Jul 19 '20 at 09:11