Gauss's Law

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Put a charge inside a small balloon, and count the field lines that come out through it. Now blow the balloon up to ten times the size, and squeeze it into a strange potato shape. Do more lines come out now, or fewer? Hold that thought for a minute. One beautiful law, Gauss's law, answers it in a single line. It comes up often in JEE, so let's begin.

0:28Gauss's law: flux through a closed surface

So, here is Gauss's law. Phi, the Greek letter, is the total flux. The integral sign with a small circle means we add over a closed surface, like our balloon. Then E vector dot d A vector; the little arrows on top mean vectors, and d A vector is a tiny piece of the surface, along the outward normal. On the right, q in, the q with the small word in below it, is the charge inside, and epsilon naught, the epsilon with a small zero below it, is the permittivity of free space. Okay? Now, back to our balloon. Blow it up or squeeze it, and the charge inside stays the same. So tell me, does the flux change? No, it does not change at all! Gauss's law holds for any closed surface, any shape, any size. Every field line from the charge has to get out through the balloon, whatever its shape or size, right? So the flux depends only on q in, the charge inside, and our puzzle is solved. Clear? Now, the E vector in this integral is the field on the surface. Is that field due only to the charge inside? No, and this is a real trap! The field on the surface is due to all the charges, inside and outside. Outside charges do change the field at each point, even though they do not change the total flux, and we will see why next. Got it?

1:50Zero flux, but not zero field

Now take a charge outside a closed surface. Every field line that enters on one side leaves on the other. Entering lines give negative flux, leaving lines give positive flux, so they cancel and phi equals zero. Right? Next, a dipole inside: plus q and minus q, together inside the surface. So q in is q minus q, which is zero, and again phi equals zero. With me? So here is the tempting step. The flux is zero, so the field on the surface must be zero too, right? No, no, wait, that is exactly the trap! Near a dipole the field is clearly not zero. Zero flux only means that the flux going in and the flux coming out cancel. So the field on the surface need not be zero. Okay?

2:39A charge and a cube: use symmetry

Now let us use this law on a cube, which comes up often in flux questions. Put a charge q at its centre. The whole charge is inside, so the total flux is q over epsilon naught. Right? Now tell me, how much of this flux goes through just one face? Yes, exactly, one sixth of it! The six faces are identical, and the charge sits right in the middle, so by symmetry each face gets q over six epsilon naught. Clear? Now move the charge to the centre of one face. Here is a topper's trick: imagine an identical twin cube on the other side of that face. The charge is now at the centre of the two cubes together, so by symmetry they share the flux equally. So our cube gets q over two epsilon naught. Right? And if the charge sits at the midpoint of an edge? Use the same trick here: how many identical cubes can meet along one edge? Four cubes, that is correct! The four cubes around that edge share the flux equally, so each one gets q over four epsilon naught. And a charge at a corner? That is our Example two. Got it?

3:48Flux through a disc, and the differential form

Now a flat disc, with a charge q on its axis. Join the charge to the rim and you get a cone; alpha, the Greek letter, is its half angle. The flux through the disc is q over two epsilon naught, times one minus cos alpha. Okay? Now let us test it. Make the disc huge, like an endless flat sheet. What happens to alpha, and to the flux? Alpha goes to ninety degrees; the arrow on screen means tends to. Cos alpha goes to zero, so the flux goes to q over two epsilon naught, exactly half. Makes sense, right? Half the lines go towards the sheet, half go away. Finally, just so you know, Gauss's law also has a differential form: del dot E vector equals rho over epsilon naught. The upside down triangle is called del, or nabla, and rho, the Greek letter, is the volume charge density. Clear?

4:44Example 1: flux of an 8.85 microcoulomb enclosed charge

Now, Example one. There is a charge of eight point eight five microcoulombs inside a closed surface; mu, the Greek letter before the capital C, means micro, ten to the power minus six. Find the flux through the surface. Does the answer depend on its shape? Pause the video and try it first. Done? Okay, let's check. Epsilon naught is eight point eight five times ten to the power minus twelve, in coulomb squared per newton metre squared; the negative powers on the units mean per. Right? So phi equals q in over epsilon naught, which is eight point eight five times ten to the power minus six, over eight point eight five times ten to the power minus twelve. Look, the eight point eight fives cancel! Very neat, isn't it? Ten to the power minus six, divided by ten to the power minus twelve, gives ten to the power six. So the flux is ten to the power six, in newton metre squared per coulomb. With me? And what about the shape? Remember our balloon from the start. The shape does not matter at all, because the flux depends only on q in. Sphere, cube or potato shape, the flux is the same. Got it?

5:53Example 2: charge at a cube corner, flux through each face

Now Example two, a little harder. This time, a charge q sits at one corner of a cube. Find the flux through each face. Pause the video and try it first. Here is a hint: add identical cubes. Done? Okay, let's check together. First, how many identical cubes can meet at one corner? Yes, exactly, eight cubes: four in the bottom layer and four on top. Together they make one big cube, with the charge q right at its centre. With me? The big cube gets q over epsilon naught, and the eight small cubes share it equally. So our cube gets one eighth, which is q over eight epsilon naught. Okay? Now, our cube has six faces, so we divide by six and get q over forty eight epsilon naught for each face, right? Oh no, careful here, because that is the trap! Look at the three faces that meet at the corner where the charge sits. The field lines from q run along these faces, never through them, so their flux is zero. Clear? So all of q over eight epsilon naught leaves through the three far faces. By symmetry each gets one third of it, which is q over twenty four epsilon naught. So, zero through the three faces at the corner, and q over twenty four epsilon naught through each far face. Got it?

7:18Quick revision: Gauss's law, zero flux, extra-cubes trick

So, quick revision. Gauss's law: the flux through any closed surface is q in over epsilon naught, whatever its shape. Zero flux does not mean zero field on the surface. And for a charge on a cube, add identical cubes until the charge sits at the centre. Now go and try the practice problems on this page. All the best!