Dielectrics and Polarisation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two charges attract each other with a force, F, in air. Now we lower them into a large tank of liquid, and also move them to half the separation. Half the distance should make the force four times stronger, right? So is the new force four times F? Hmm, the liquid changes things. Hold that thought. Dielectrics come up often in JEE problems. So, let's begin.

0:28Two kinds of molecules

So, what is inside a dielectric? Molecules, and they come in two kinds. First, the non-polar ones, like oxygen, nitrogen and carbon dioxide, written O two, N two and C O two. On their own, they have no permanent dipole moment. Okay? Now let me ask you this. Put one of these molecules in an electric field. The field pushes its positive charge one way and its negative charge the other way. So what does the molecule become? Yes, exactly, it becomes a dipole, an induced dipole. The field itself induces a dipole moment that was not there before. Clear? Second, the polar molecules, like water and hydrogen chloride, written H two O and HCL on screen. These already have permanent dipoles, even with no field at all. Right? Switch on a field, and these dipoles turn to line up with it, but only partly. Either way, the material ends up with dipoles pointing along the field. This is what we call polarisation. With me?

1:32Bound charge weakens the field

Now picture a slab of dielectric in a field. The lined-up dipoles leave a thin layer of charge on the two faces of the slab, called bound charge. And here is the key point: the bound charge opposes the applied field. Got it? So the field inside becomes weaker: E equals E naught over K, where E naught, the E with a small zero, is the applied field, and K is the dielectric constant of the material. Okay? Now think about this. The bound charge always opposes the applied field. So can the dielectric constant ever be less than one? No, it cannot, and the new line says so. K equals epsilon r, which equals epsilon over epsilon naught, and it is greater than or equal to one. Epsilon r, the Greek epsilon with a small r, is the relative permittivity; epsilon is the permittivity of the material, and epsilon naught, the epsilon with a small zero, is that of free space. Right? And at the other extreme, a metal. For a metal, K tends to infinity; that is the arrow and the infinity sign. Then E naught over K goes to zero, which matches what we know: the field inside a conductor in electrostatic equilibrium is zero. Clear?

2:47How much bound charge?

Now, how much bound charge is there? Take a slab facing a free charge q, like the charge on a capacitor plate. The bound charge on the face next to it is q b, the q with a small b, and q b equals q times, in brackets, one minus one over K, opposite in sign to q. Okay? Let us test this formula. For a metal, K tends to infinity. So what does q b become? One over K goes to zero, so q b becomes q itself, perfect! The bound charge cancels the free charge completely, and the field inside is zero. Everything fits, right? And here is a topper's trick to remember it. The free charge q, minus the bound charge q b, leaves q over K, and that left-over charge is exactly why the field falls to E naught over K, got it?

3:39Filled capacitors and charges in a medium

Now let us come to the capacitor. Fill the gap completely with a dielectric, and C equals K times C naught, where C naught, the C with a small zero, is the capacitance with air in the gap. For the same charge, the field and the voltage both fall by K, so the capacitance goes up K times. Okay? But remember, partly filled capacitors need a different formula, which we take up separately. Right? And for point charges in a dielectric medium, Coulomb's law is simply divided by K, so F equals one over four pi epsilon naught K, times q one q two over r squared. Here q one and q two are the charges, r is the distance between them. The medium must surround the charges, like a large tank of liquid. With me? So the rule is simple: in a medium, forces and fields are both reduced by K, and our puzzle needs exactly this. Clear?

4:32Dielectric strength

Now one more idea. Every material has a limit, called its dielectric strength: the largest field it can withstand without breakdown. Okay? For air, it is about three times ten to the power six volt per metre; V is volt, and m to the power minus one means per metre. Push the field beyond this, and the air breaks down. Got it?

4:53Example 1: bound charge and field in a capacitor with K = 5

Here is the first example: a capacitor has plus or minus ten microcoulomb on its plates, so one plate has plus ten and the other minus ten; mu, the Greek letter, means micro. It is filled with a dielectric of K equals five. Find the bound charge on each face, and the field inside compared with that without it. Pause the video and try it first. Done? Okay, let's check. Many students think like this: the dielectric constant is five, so ten over five gives two microcoulomb. Is that the bound charge? No, no, wait, that is not the bound charge. Use the formula: q b equals ten times, in brackets, one minus one fifth. That is ten times four fifths, which gives eight microcoulomb. Okay? Now let us look at the signs. The face next to the plus ten plate gets minus eight microcoulomb, opposite in sign, and the other face gets plus eight. Clear? And what about the field? The plate and the face next to it together carry ten minus eight, which is two microcoulomb, one fifth of ten. Right? So the field is E naught over five, one fifth of its value with air. See, the two microcoulomb from our trap is the left-over charge that sets the field, not the bound charge. Got it?

6:13Example 2: force in a liquid of K = 4 at half the distance

And now, our puzzle from the start. Two charges attract with a force, F, in air. They are placed in a large tank of liquid of K equals four, and moved to half the separation. Find the force. Pause the video and try it first. Done? Let's check. The tank is large, so the liquid surrounds both charges, and F is proportional to one over K times r squared; the sign on screen that looks like an open alpha means proportional to. Okay? Take the liquid first: K is four, so the force is multiplied by one quarter. So the answer is F over four, right? Oh no, careful, we forgot the distance! The separation r becomes r over two; the arrow means becomes. Then one over r squared becomes four times bigger, a factor of four. With me? So the new force is F times one quarter times four, which is just F! So the answer to our puzzle is not four times F, but exactly F, the same as in air. The liquid takes away exactly what the shorter distance gave. Isn't that neat?

7:21Quick revision: E0/K, bound charge, C = KC0, force in medium

So, quick revision. First, bound charge opposes the field, so E equals E naught over K, with K at least one. Second, the bound charge is q times, one minus one over K, opposite in sign, and a completely filled capacitor has C equals K times C naught. And third, in a medium around the charges, forces and fields are reduced by K, okay? Now go and try the practice problems on this page. All the best!