Partially Filled Capacitors

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Slide a dielectric slab into the gap of a parallel plate capacitor, covering the whole plate area, but thinner than the gap. Does it matter where the slab sits, touching one plate, or right in the middle? Today one simple idea will settle it. Partially filled capacitors come up often in JEE, so let's begin.

0:26Layers across the gap act in series

First, a quick reminder. With only air between the plates, the capacitance is C naught, the C with a small zero, equal to epsilon naught times capital A, over d. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space; capital A is the plate area, and d is the gap. Okay? Now look at the left picture. The dark lines are the plates, with a slab of thickness t and dielectric constant capital K between them, and an air gap on each side. We cross these layers one after another. Let me ask you this: do they act in series, or in parallel? Yes, exactly, in series, like capacitors in a chain! Filling a gap completely multiplies the capacitance by K, so layer i, of thickness t i and constant K i, is a capacitor C i, equal to K i times epsilon naught times capital A, over t i. The small i is a label for the layer. Right? In series, the reciprocals add. So one over C is the sum of t i, over K i times epsilon naught times capital A; the big Greek letter sigma means add over all the layers. Clear?

1:38One slab across the whole area

Now the most common case, a single slab of thickness t, covering the whole plate area. The two air gaps add up to d minus t. So one over C equals d minus t, over epsilon naught times capital A, plus t, over K times epsilon naught times capital A, right? Take epsilon naught times capital A common, and flip it. C equals epsilon naught times capital A, over d minus t, plus t over K, the key formula of today, also written under the left picture. Okay? So let me ask you this: compared with C naught, does the slab raise the capacitance, or lower it? Yes, the slab raises it, or at worst leaves it the same. K is greater than or equal to one, the sign with a line under it, so t over K is at most t, and C is at least C naught. Oh no, careful here: writing t times K would lower the capacitance, so that cannot be right. Got it?

2:34Where the slab sits, and a metal slab

Now, back to our puzzle. Look at the formula under the left picture once more. Does it matter where the slab sits in the gap? No, it does not matter at all! Only the total air gap, d minus t, enters the formula. Move the slab, and one gap shrinks as the other grows by the same amount. So the capacitance stays the same, and our puzzle is solved. Clear? Now, what if the slab is metal? For a metal, capital K tends to infinity; the arrow means tends to, and the sideways eight is infinity. So t over K tends to zero, okay? So that term drops out, and C equals epsilon naught times capital A, over d minus t. The metal slab simply cuts the gap down to d minus t. Right?

3:25Side by side: parallel

Now turn the materials the other way, as in the middle picture. K one and K two, with small labels one and two, sit side by side, and each runs from one plate to the other. So tell me, series or parallel this time? Yes, this time they are in parallel! Both pieces touch the same two plates, so they share the same voltage. Clear? So the capacitances add. Each piece, of constant K i and area capital A i, gives K i times epsilon naught times capital A i, over d, so C equals epsilon naught over d, times the sum of K i times capital A i, just as written under the middle picture. Got it? The right picture mixes both ideas: K one and K two are in parallel, shown by the two vertical bars, and that pair is in series with K three. Right?

4:13Permittivity that changes across the gap

Now one last case, which comes up in harder problems. Sometimes the permittivity changes as we cross the gap. We write it as epsilon of x, where x is the distance from one plate. Okay? So how do we handle it? We use our layer idea again! Cut the gap into very thin layers of thickness d x, stacked across the gap, so they are all in series. With me? Each thin layer adds d x, over epsilon of x times capital A, to one over C, and adding infinitely many of them is an integral. So one over C equals one over capital A, times the integral from zero to d of d x over epsilon of x; the long S is the integral sign. Clear?

4:56Example 1: slab of K = 4 in half the gap, new capacitance

Now our first example. We place a slab, with K equal to four and thickness d over two, in a capacitor C naught of gap d. Find the new capacitance. Pause the video and try it first. Done? Okay, let's check. Many students write four times C naught, because K is four. So, is that right? No, no, careful, four times C naught needs the slab to fill the whole gap, and here it fills only half! So, d minus t is d over two, and t over K is d over two, divided by four, which is d over eight. Together they give five d over eight. Okay? So C equals epsilon naught times capital A, divided by five d over eight, which is eight over five, times epsilon naught times capital A, over d; the small dot means times. That is eight fifths of C naught, or one point six times C naught. Got it?

5:51Example 2: isolated capacitor, slab of K = 3, new voltage

Now a harder one. Here a capacitor is charged to V naught, the V with a small zero, and isolated. Then a slab of thickness t equal to d over two, with K equal to three, slides in. Find the potential difference now. Pause the video and try it first. Done? Okay, let's check. First, the key question: once the capacitor is isolated, what stays fixed, the charge or the potential difference? Yes, exactly, the charge, capital Q, stays fixed! There is no battery, so it has nowhere to go. With the battery connected, the voltage would stay fixed instead, and that is a different question. Clear? So V naught is Q over C naught, or Q times d, over epsilon naught times capital A, and the new V is Q over the new C, or Q times, in brackets, d minus t plus t over K, all over epsilon naught times capital A, right? Divide the two, and epsilon naught times capital A cancels. So V equals V naught, times d minus t plus t over K, over d. With me? Now put t equal to d over two, and K equal to three. d minus t, over d, is one half, and t over K, divided by d, is one sixth. So V is V naught times, in brackets, one half plus one sixth, which gives two thirds of V naught. Got it?

7:15Quick revision: series, parallel, slab and metal slab

So, quick revision. Layers across the gap act in series, and pieces side by side act in parallel. One slab gives C equals epsilon naught times capital A, over d minus t plus t over K, wherever it sits; for metal, drop the t over K term. And an isolated capacitor keeps its charge, so V equals Q over C, okay? You now have the key ideas of this topic. Now go and try the practice problems on this page. All the best!