Inserting a Dielectric: Battery Connected or Disconnected

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take a charged capacitor, and slide a slab of glass into the gap between its plates. Does the stored energy go up, or go down? Many students answer too quickly, because the answer depends on one simple question: is the battery still connected? This comes up often in JEE, so let's settle it today.

0:25First ask: what stays fixed?

So, before any formula, one question. We charge a capacitor, then disconnect the battery, so the plates are joined to nothing. Tell me, can the charge on them change? No, it cannot. There is no wire to bring charge in or take it out. So with the battery disconnected, Q, the charge on the plates, stays fixed. Okay? Now keep the battery connected instead. The battery holds the plates at its own voltage, so V, the potential difference, stays fixed. Right? In both cases, the slab fills the whole gap, and K is its dielectric constant. The arrow on screen means becomes, so the capacitance C becomes K times C, okay? Always ask what stays fixed before writing anything else.

1:13Battery disconnected: Q fixed

Case one: the battery is disconnected. So Q is fixed, and the capacitance becomes K times bigger. Okay? Now you tell me. Q stays the same, while C becomes K times bigger. So what happens to V, the potential difference? It falls, yes, exactly! V equals Q over C, so V becomes V over K: it is K times smaller. And E, the electric field between the plates, falls by the same factor, to E over K, okay? Now U, the stored energy. The capacitance has become K times bigger, so the energy also becomes K times U, right? No, wait, that is the trap, because V did not stay the same. U is one half C V squared. C went up K times, but V went down K times, and V is squared. So one half, times K C, times V over K in brackets, squared, gives U over K, and the energy falls. Got it? So where did the lost energy go? The field pulls the slab into the gap, and does work on it. That work is U minus U over K, which is U times, one minus one over K, in brackets. With me?

2:23Battery connected: V fixed

Now case two: the battery stays connected. V is fixed, and again the capacitance becomes K times bigger. The field E is V over d, where d is the gap between the plates; both stay the same, so the field does not change. Okay? Your turn now. V is fixed, and C grows K times. So what happens to the charge on the plates? It grows K times, that is correct! Q equals C times V, so Q becomes K times Q: the battery pushes extra charge onto the plates. Right? And the energy, one half C V squared? V is fixed and C is K times bigger, so this time the energy does rise, to K times U, okay? Same slab, opposite answer, and that solves our puzzle from the start. Clear? But now a deeper question. The stored energy went up. So do we have to push the slab in, against the field? No, the slab is still pulled in, and here is why. Take Q as the charge before the slab went in. The battery pushes the extra charge, K minus one in brackets, times Q, through the voltage V, so the energy it supplies is K minus one, times Q times V, okay? Only half of it is stored, and the other half is the work done on the slab, pulling it in.

3:43The pull on a slab, and a rising liquid

So how strong is this pull? Keep V fixed, and push a slab of width b, as thick as the gap d, partly into the capacitor. The force pulling it in is epsilon naught, times b, times K minus one, times V squared, over two times d. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Okay? And look, the distance x, how far the slab is in, does not appear at all. So the pull is the same, wherever the slab is. Right? The same pull lifts a liquid. Dip two vertical plates into a liquid with dielectric constant K and density rho, the Greek letter that looks like a p, and keep a voltage V between them. The liquid rises to a height h, equal to epsilon naught, times K minus one, times V squared, over two times rho times g times d squared, with g the acceleration due to gravity. Clear?

4:40Example 1: isolated 10 micro F at 100 V, K = 5 slab fills it

Now, Example one, from the course. Ten microfarad capacitor, charged to one hundred volts, and then isolated. The Greek mu on screen means micro. Then a slab with K equal to five fills it completely. Find the new potential difference, the new energy, and the work done by the field on the slab. Pause the video and try it first. Done? Okay, let's check. Isolated means the battery is gone, so Q is fixed, and the voltage falls K times. One hundred over five gives twenty volts. Okay? Now the starting energy, U nought, with a small zero. One half C V squared is one half, times ten to the power minus five, the capacitance in farad, times ten to the power four, which is one hundred squared. That gives zero point zero five joule. Right? With Q fixed, the energy falls to U nought over five. Zero point zero five over five gives zero point zero one joule. Clear? And the work done by the field on the slab is exactly the energy lost. Zero point zero five minus zero point zero one gives zero point zero four joule, and that's it!

5:52Example 2: same slab with the 100 V battery kept connected

Now Example two, a little harder. Same capacitor, same slab, but this time the one hundred volt battery stays connected. Find the new charge, the new energy, and the energy supplied by the battery. Pause the video and try it first. Done? Okay, let's check. The battery is connected, so V is fixed, and the charge becomes K times bigger. At the start, the charge is ten microfarad times one hundred volts, one millicoulomb, the m C on screen. Five times that gives five millicoulomb. Okay? The energy becomes K times U, so zero point zero five joule becomes zero point two five joule. It rose this time, five times bigger. Right? Now the battery. The stored energy went up by zero point two joule. So the battery supplied zero point two joule, right? No, careful, that is the trap. The battery pushed extra charge, delta Q, through one hundred volts. The Greek delta, the little triangle, means change in. Delta Q is five minus one, four millicoulomb, which is four times ten to the power minus three coulomb. Times one hundred, that gives zero point four joule, twice the rise in stored energy. Clear? So, zero point two joule is stored, and the other zero point two joule is the work done on the slab. Half and half, just as K minus one, times Q times V, told us. And compare with Example one: the slab was pulled in both times, but the energy fell once and rose once. That's it!

7:30Quick revision: Q fixed or V fixed, and the battery energy

So, quick revision. First, ask what stays fixed: the charge when the battery is disconnected, the voltage when it is connected. With the charge fixed, V, E and U all become K times smaller. With the voltage fixed, Q and U rise K times, and the field stays the same. And when connected, the battery gives K minus one, times Q times V, with Q the charge before the slab went in: half is stored, half pulls the slab in. Now go and try the practice problems on this page. All the best!