Spheres, Shells and Cavities

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Take a ball with charge spread evenly through it, and scoop out a round hole, away from the centre. Inside a hollow charged shell, the field is zero. So inside this hole, it is zero too, right? Hmm, hold that thought. By the end, you will settle it in two lines. This comes up often in JEE, so let's begin.

0:26Thin shell: zero inside, a jump at the surface

First, a thin shell of radius capital R, with charge capital Q spread evenly over it. Look at the first panel of the figure: a dashed Gaussian sphere, drawn there around a point charge. Ours has radius small r and the same centre as the shell. Gauss's law needs a closed surface like this one. By symmetry, the field E has the same size all over it and points along the radius, so the flux is E times four pi r squared, the dot meaning times. Gauss's law sets this equal to q in, the charge enclosed, with the small word in written below the q, over epsilon naught, the Greek letter with a small zero. Clear? Now shrink the Gaussian sphere to sit inside the shell. How much charge does it enclose? There is none at all, exactly! For r less than capital R, all the charge is on the shell, outside our sphere. So q in is zero, and E is zero everywhere inside. In the graph in the last panel, that is the flat red line. Got it? Outside, the Gaussian sphere encloses all of capital Q, so E equals k Q over r squared, with k equal to one over four pi epsilon naught. That is the falling curve in the graph. Right? Just outside, at r equal to capital R, E is k Q over capital R squared, which equals sigma over epsilon naught, where sigma, a Greek letter, is the charge per unit area. So E jumps from zero to sigma over epsilon naught at the surface, the red dashed line. With me?

1:57Uniform solid sphere: grows inside, no jump

Now fill the ball completely: a solid sphere, with capital Q spread evenly through its volume. Inside the shell, the field was zero. So inside the solid sphere, it is zero too, right? Oh no, careful, that is a common trap! Now a Gaussian sphere inside does enclose charge: the same fraction of capital Q as its fraction of the volume. So q in equals capital Q times r cubed over capital R cubed. Okay? Gauss's law then gives E inside equal to k Q r over capital R cubed, or rho times r over three epsilon naught, where rho, a Greek letter, is the charge per unit volume. It grows with r, the straight line marked solid. Clear? Outside, all of capital Q is enclosed, so E is k Q over r squared again. So where is the solid sphere's field the largest: at the centre, at the surface, or far away? It is at the surface, yes! At r equal to capital R, both formulas give k Q over capital R squared. So the field peaks at the surface with no jump, unlike the shell. With me?

3:07Non-uniform density: add up thin shells

Now, what if the density is not uniform? Say rho equals rho naught times r to the power n, with rho naught, rho with a small zero, a constant. Can we still use capital Q times r cubed over capital R cubed? We cannot, since that fraction needs a uniform density. So we add up thin shells of radius r prime, the r with a dash. Each shell carries its density, rho naught times r prime to the power n, times its volume, four pi r prime squared, d r prime. Integrating from zero to r, the long S sign, q of r is four pi rho naught, r to the power n plus three, over n plus three. Okay? Then Gauss's law gives E equals q of r over four pi epsilon naught r squared, which is rho naught times r to the power n plus one, over the bracket n plus three, times epsilon naught. Clear?

4:00Topper's trick: fill the cavity

Now, a topper's trick for our puzzle. Fill the hole with plus rho and minus rho together; they cancel, so nothing changes. But now we have two simple spheres: a full sphere of plus rho, and a cavity-sized sphere of minus rho. Okay? So is the field inside the cavity zero, like inside a shell? No, and this surprises many students: it is not zero! E vector equals rho times a vector, over three epsilon naught, where the arrows mean vectors, and a vector goes from the centre O to the cavity's centre. It is uniform, and example two proves it. Got it? And when spheres of plus rho and minus rho overlap, the overlap has a uniform field, rho times d vector over three epsilon naught, with d vector from the plus centre to the minus centre. Clear?

4:49Example 1: E at 5, 10, 20 cm from a charged solid sphere

Let's try our first example. Here is a uniformly charged solid sphere of radius ten centimetres and charge one microcoulomb, where the Greek letter mu means micro. Find E at five, ten and twenty centimetres from the centre. Pause the video and try it first. Done? Okay, let's check. With k as nine times ten to the power nine, and capital Q as one times ten to the power minus six, k Q is nine times ten to the power three. And capital R is zero point one metre. Right? Now, five centimetres. Put it into k Q over r squared, and you get three point six times ten to the power six, right? Oh no, careful, that is the trap! Five centimetres is inside, so use k Q r over capital R cubed: nine times ten to the power three, times zero point zero five, over zero point one cubed. That gives four point five times ten to the power five newton per coulomb, written N C to the power minus one. Got it? Ten centimetres is the surface, so k Q over capital R squared. Nine times ten to the power three over zero point one squared gives nine times ten to the power five newton per coulomb, the largest value. Okay? Twenty centimetres is outside, so k Q over r squared. Nine times ten to the power three over zero point two squared gives two point two five times ten to the power five newton per coulomb. Clear?

6:17Example 2: field inside an off-centre spherical cavity

Now the harder one, and it settles our puzzle. Here is a sphere of radius capital R and uniform density rho, except for a spherical cavity of radius capital R over two, whose centre is capital R over two from O, the centre of the sphere. Find the field inside the cavity. Pause the video and try it first. Done? Okay, let's check. Fill the cavity with plus rho and minus rho, giving a full sphere of plus rho centred at O, and a small sphere of minus rho. Okay? Take any point in the cavity, at r vector from O, and so at r vector minus a vector from the cavity centre. With me? Now, for each sphere, do we use the inside formula or the outside formula? Inside for both, exactly, as the point is inside both spheres. So E vector is rho times r vector over three epsilon naught, minus rho times the bracket r vector minus a vector, over three epsilon naught. The r vector cancels, leaving rho times a vector, over three epsilon naught. Right? The modulus of a vector, shown by the two vertical bars, is capital R over two. So E equals rho times capital R, over six epsilon naught, along a vector. So our puzzle is solved: the field in the hole is uniform, not zero. Clear?

7:39Quick revision: shell, solid sphere and the cavity trick

So, quick revision. For a thin shell, zero inside, k Q over r squared outside, and a jump at the surface. For a uniform solid sphere, growing with r inside, largest at the surface, no jump. For a cavity, add plus rho and minus rho, and get a uniform field, rho times a vector over three epsilon naught. Now try the practice problems on this page. All the best!