Concentric Shells, Earthing and Connections

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a charged metal shell and connect it to the earth with a wire. Does all its charge run away to the earth? The quick answer is yes, right? Hold that thought, because by the end you will see that it is not always so. Shells, earthing and wires come up often in JEE questions. So, let's begin.

0:25Potential of concentric shells

So, picture thin metal shells, one inside the other, with the same centre: concentric shells. Shell number j carries charge q j, spread evenly over radius capital R j; the small j below is just a label. And k is one over four pi epsilon naught, with epsilon naught the permittivity of free space. Now take a point at distance r from the centre. If a shell lies inside this point, closer to the centre, then capital R j is less than r; that is the less-than sign. Then the shell acts like a point charge at the centre, giving k q j over r. Clear? Now the other case: the point is inside the shell. What potential does the shell give there? It is a constant, yes, exactly! Inside a shell, its own field is zero, so its potential stays at its surface value, k q j over capital R j; the greater-than sign says capital R j is bigger than r. With me? Now both cases in one line. V of r, the potential at distance r, equals k times the sum of q j over the max of r and capital R j. The big Greek letter sigma means add over all shells, and max means the bigger of the two. Potential is a scalar, so no arrows: just add, with signs. Okay?

1:44Charges on the surfaces of a conducting shell

Our shells are metal, so they are conductors. In the left part of the figure, a sphere with charge plus q sits inside a thick metal shell, the shaded ring, and capital Q is the shell's own charge. In electrostatic equilibrium, the field E inside the metal is zero. Right? So tell me, what charge must sit on the inner surface of this shell? Minus q, yes, exactly! See the little minus signs on the inner surface. The rule is that the inner surface of a shell carries minus the total charge enclosed by it. Got it? Here is why. Take a closed Gaussian surface inside the metal, around the cavity. The field is zero on it, so the flux is zero, and by Gauss's law the enclosed charge is zero. So the inner surface must cancel the plus q. Clear? And the outer surface carries the rest. The shell's total is capital Q, with minus q already inside, so the outer surface gets capital Q plus q, the little plus signs. With me?

2:49Earthing and joining by a wire

Now, earthing: we connect a conductor to the earth with a wire. In the right part of the figure, the three little lines are the earth symbol. Charge can now flow to the earth, or from it. Okay? So, after earthing, all the charge runs away and the shell has zero charge, right? No, no, careful, that is the trap from our puzzle! Earthing makes the potential V zero, not necessarily the charge q. Got it? So an earthed conductor keeps whatever charge makes its potential zero. In the figure, the inner shell has capital Q, and the earthed outer shell keeps minus capital Q, not zero. Right? Next, a wire between two conductors: charge flows until their potentials are equal. Now, what if the wire joins an inner shell to the outer shell around it? Where does the charge go? All of it goes to the outer surface of the outer shell, exactly! Joined, the two shells are at the same potential, so no field can remain between them, and then the inner shell can hold no charge. With me?

3:56Topper's trick: the potential difference

Now, a topper's trick. Take an inner sphere of radius small a inside a concentric shell of radius b, with potentials V a and V b; the small a and b below are labels. Now, does the charge on the outer shell change V a minus V b? No, it does not change it at all! The outer shell adds the same constant to both potentials, so it cancels in the difference. V a minus V b equals k q in, times one over small a minus one over b; the small word in, below q, means the inner charge. Okay? In the right part of the figure, the outer shell is earthed, so V b is zero. So V inner, with inner written small below, equals k times capital Q, times one over small a minus one over b. Quick, right?

4:44Example 1: potentials of shells R and 2R with q and -q

Now, our first example. Two concentric shells, of radii capital R and two R, carry charges q and minus q. Find the potential of each shell. Pause the video and try it first. Done? Okay, let's check. At the inner shell, r equals capital R, and V with a small R below is its potential. The inner shell gives k q over R, and the outer shell, radius two R, is outside this point, so it gives its surface value, k times minus q over two R, okay? So V R equals k q over R, minus k q over two R, which gives k q over two R, right? Now the outer shell, where r equals two R: both shells are inside or on this point, so both act like point charges. What do you get? It comes out to zero, yes, exactly! V two R equals k times q minus q, over two R, and q minus q is zero, so V two R is zero. With me? Now check with the topper's trick: k q, times one over R minus one over two R, gives k q over two R, and since V two R is zero, this matches V R exactly. Perfect, both methods agree!

5:58Example 2: new charge on the earthed middle of three shells

Now a harder one. Three concentric shells, of radii small a, two a and three a, carry charges q, two q and minus three q. The middle shell is earthed. Find its new charge. Pause the video and try it first. Done? Okay, let's check. The middle shell is earthed, so its charge becomes zero, right? No, no, careful, that is our puzzle again! Earthing makes the potential zero, not the charge. So call the new charge q prime; the little dash is a prime, just a new name, and we find it from V equals zero. Okay? V two a, with the small two a below, is the potential of the middle shell. The inner shell and the middle shell are inside or on this point, so together they give k times q plus q prime, over two a. The outer shell, minus three q, is outside, so it gives its surface value, minus k times three q over three a. The total equals zero. With me? Three q over three a is just q over small a. So q plus q prime, over two a, equals q over small a; the double arrow means which gives. Multiply both sides by two a, and q plus q prime equals two q. Clear? So q prime equals q. The middle shell started with two q and keeps q, so it loses q to the earth, and its charge is not zero at all! Earthing fixed the potential, not the charge, and that answers our puzzle. Got it?

7:29Quick revision: shell potential, earthing, connecting wire

So, quick revision. For concentric shells, V of r is k times the sum of q j over the bigger of r and capital R j. Earthed means the potential is zero, not the charge. And a wire means equal potentials; joining an inner shell to the outer one sends all the charge outside. Now try the practice problems on this page. All the best!