Lines, Cylinders, Sheets and Slabs

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Walk twice as far from a point charge, and its field drops to a quarter, right? Now walk twice as far from a huge, flat sheet of charge. What happens to the field this time? Keep your answer; Gauss's law will settle it. Lines, sheets and slabs come up often in JEE, so let's begin.

0:24Long line charge: a Gaussian cylinder

First, a long, straight line of charge. In the figure, parts a and d are the sphere cases from before; we need b and c. In part b, lambda, the Greek letter at the top, is the linear charge density, charge per unit length. Around the line we draw a dashed Gaussian cylinder of radius r and length l. Okay? By symmetry, the electric field E points straight out from the line, the same all around. So, through which parts of this cylinder does the flux pass? Only through the curved side, exactly! The field runs along the flat ends, so no flux there. So E times two pi r times l equals the charge inside, lambda times l, over epsilon naught. Epsilon naught, the epsilon with a small zero below it, is the permittivity of free space, and the dot means times. Right? Cancel l, and E equals lambda over two pi epsilon naught r. So a line's field falls as one over r, not one over r squared; the sign like an open eight lying on its side means proportional to. Strictly, this holds for an infinitely long line. Clear?

1:36Long cylinder of charge, and a cylindrical shell

Next, a long solid cylinder, uniformly charged. Rho, the Greek letter, is the volume charge density; capital R is the radius, and small r is our distance from the axis. Draw a Gaussian cylinder of radius r through a point inside. Does it enclose all the charge? No, only the charge within radius r. So E equals rho times r, over two epsilon naught, growing with r. With me? Outside, it encloses all the charge, so E equals rho times capital R squared, over two epsilon naught r. It falls as one over r, just like a line. Got it? And a charged cylindrical shell, a hollow pipe with charge only on its surface? Any Gaussian cylinder inside it encloses no charge, so the field is zero. Right?

2:25Infinite sheet: a pillbox

Now, back to our puzzle, the flat sheet. Sigma, the Greek letter in part c, is the surface charge density. Our Gaussian surface is a pillbox, a short cylinder through the sheet, with capital A the area of each face. The field is perpendicular to the sheet, so both flat faces carry flux and the curved side carries none, and two E times A equals sigma times A, over epsilon naught. Okay? Solve it for the field. Does the distance from the sheet appear anywhere? No, it does not appear at all! Capital A cancels, and E equals sigma over two epsilon naught on each side. Twice as far, ten times as far, the field of an infinite sheet stays the same. Puzzle solved: as long as the sheet is huge compared with your distance from it, the field does not change at all. Clear?

3:17Two or more sheets: add with directions

Now, what about several sheets? Each gives sigma i over two epsilon naught; the small i below sigma numbers the sheets. It points away from a positive sheet and towards a negative one, and we add the fields with their directions. With me? Take two parallel sheets, plus sigma and minus sigma. Between them, and outside them, do the fields add or cancel? Between them, both fields point from plus to minus, so they add, giving sigma over epsilon naught. Outside, they point opposite ways and cancel, so the field is zero. Okay? Now make both sheets positive. So between them it is again sigma over epsilon naught, right? No, wait, that is exactly the trap! Between two positive sheets, the fields point opposite ways and cancel, giving zero. Outside, they point the same way, giving sigma over epsilon naught. Got it?

4:16Thick slab, and just outside a conductor

Next, a thick slab of thickness two d and uniform charge density rho. We measure x from the mid-plane, the plane right in the middle. Right? Inside, a pillbox from minus x to plus x gives E equals rho times x, over epsilon naught. Once x reaches d, the enclosed charge stops growing, so outside, E equals rho times d, over epsilon naught, constant, like a sheet. Clear? Now, just outside a charged conductor, the surface looks like a sheet of density sigma. So the field there is sigma over two epsilon naught, right? Oh no, careful, that is a classic trap! Inside a conductor in electrostatic equilibrium, the field is zero, so the pillbox has flux only through its outer face. So E equals sigma over epsilon naught, normal to the surface, twice the sheet value. With me?

5:12Example 1: charge density of a wire with E known at 2 cm

Now, Example one. Here, a long wire produces nine times ten to the power four newton per coulomb at two centimetres; capital N is newton, capital C is coulomb, and the power minus one means per. Find its linear charge density. Pause the video and try it first. Done? Okay, let's check. Here k, one over four pi epsilon naught, is nine times ten to the power nine, in newton metre squared per coulomb squared. From the line formula, lambda equals two pi epsilon naught times r times E, which is r times E, over two k. Okay? The distance is two centimetres, so we put r equals two, right? No, careful here, because the formula needs metres. Two centimetres is zero point zero two metre. Got it? Two times k is one point eight times ten to the power ten. So lambda equals zero point zero two times nine times ten to the power four, over one point eight times ten to the power ten. That gives ten to the power minus seven coulomb per metre. With r equals two, you would be a hundred times too big. Clear?

6:23Example 2: three parallel sheets, field in all four regions

Now Example two, a little harder. Three large parallel sheets with densities sigma, two sigma and minus sigma sit at x equals zero, x equals a, and x equals two a. Find the field in all four regions. Pause the video and try it first; draw an arrow for each sheet. Done? Okay, let's check. Here is a trick that toppers use: count in units of sigma over two epsilon naught. Then the sheets give one, two and one unit, and the plus x direction is positive. Okay? First, left of all three sheets, where x is less than zero. Which way does each field point here? The two positive sheets' fields point away, to the left, so minus one and minus two. The negative sheet's field points towards it, to the right, so plus one. The total is minus two units, so E equals minus sigma over epsilon naught. Right? Between zero and a, the first sheet is behind us, so its field points right, plus one. The others still give minus two and plus one. One minus two plus one is zero, so no field here at all. Clear? Between a and two a, all three fields point right, plus one, plus two and plus one. That is four units, so E equals two sigma over epsilon naught. With me? Beyond two a, the negative sheet is behind us, so its field points left. One plus two minus one gives two units, so E equals sigma over epsilon naught, along plus x. Got it?

7:58Quick revision: line, sheet, several sheets, conductor

So, quick revision. For a line, E equals lambda over two pi epsilon naught r, falling as one over r. For a sheet, sigma over two epsilon naught, at every distance. For several sheets, add with directions, and just outside a conductor, it is sigma over epsilon naught. Now go and try the practice problems on this page. All the best!