Conductors in Electrostatic Equilibrium

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two metal spheres far apart, one small and one big, are joined by a thin wire and given some charge. The big sphere ends up with more of it. So which sphere has the stronger electric field at its surface? The big one, surely, right? Hmm, hold that thought. Conductors come up often in JEE electrostatics, and this one has a surprise in it. So, let's begin.

0:29Inside a conductor: the field is zero

So, take a metal conductor in electrostatic equilibrium, which means its charges have settled down, okay? Every metal has free electrons. So let me ask you this: can there be an electric field inside the metal itself? The answer is no. Any field would push those free electrons, they would move, and that is not equilibrium. So inside the material, E, the electric field, is zero. Right? With zero field, no work is done in moving a charge from one point of the metal to another. So the potential, capital V, is the same everywhere, and the whole conductor is an equipotential. Clear? Next, where does any extra charge go? Take a closed Gaussian surface just inside the metal. The field on it is zero, so by Gauss's law, it encloses no charge. So any net charge lies on the outer surface. With me? Just outside, the field E is sigma over epsilon naught, times n hat. Sigma is the surface charge density, the charge per unit area, and epsilon naught, the epsilon with a small zero, is the permittivity of free space. The arrow on E marks a vector, and n hat, the n with a little cap, is the unit vector normal to the surface, pointing outwards. Okay?

1:45A cavity inside the conductor

Now cut a hollow, a cavity, inside the conductor. If the cavity is empty, there is no field inside it, and no charge on its wall. The metal shields the cavity, and we call this electrostatic shielding. Got it? Now place a point charge q inside the cavity. What charge appears on the cavity wall? Yes, it is minus q. Draw a closed Gaussian surface inside the metal, around the cavity. The field on it is zero, so it encloses zero charge, and the wall must carry minus q. And a neutral conductor stays neutral, so plus q appears on the outer surface. Right? If the conductor has a charge of its own, that also sits on the outer surface, along with the plus q. Okay? And for a spherical conductor, put q anywhere in the cavity, even off the centre, and the outer charge still spreads uniformly. Clear?

2:40Sharp points, and spheres joined by a wire

Now, the charge on the outer surface is not spread evenly, in general. Sigma is larger where the surface is sharper. With me? To see this, take two spheres far apart, of radii R one and R two, joined by a thin wire. They now form one conductor. So what must be the same for both? Yes, exactly, the potential, because a conductor is an equipotential! So k q one over R one equals k q two over R two, where k is one over four pi epsilon naught. The small one and two below are just labels. Okay? Cancel k, and q one over q two equals R one over R two. The bigger sphere takes the bigger charge. Right? But sigma one over sigma two equals E one over E two, which equals R two over R one, the ratio flipped! So the smaller, sharper sphere has the larger sigma and the stronger field. Got it?

3:36Outward pressure on a charged surface

Now, the surface charges repel each other, so the surface is pushed outwards. The field just outside is sigma over epsilon naught, and just inside, it is zero, okay? So the force per unit area is sigma times sigma over epsilon naught, right? Oh no, careful, this is exactly where students slip! The surface charge feels the average of the outside and inside fields, so sigma over epsilon naught plus zero, over two, which is sigma over two epsilon naught. Clear? So the outward pressure is sigma times sigma over two epsilon naught, that is, sigma squared over two epsilon naught. Remember the two at the bottom. With me?

4:16Identical drops merging into one

Now let us look at one last situation. Take n identical drops, each with charge q, potential small v and radius small r, and merge them into one big drop. Okay? The total volume stays the same, so the big radius, capital R, is n to the power one third, times r, the cube root of n, times r. Right? The charge is now n times q, so the potential, capital V, is k times n q, over capital R, which works out to n to the power two thirds, times small v. Clear? In the same way, sigma prime, the surface charge density of the big drop, is n to the power one third, times sigma, and U prime, its energy, is n to the power five thirds, times small u, the energy of one small drop. The prime, that small dash, marks the big drop. Got it?

5:08Example 1: 2 cm and 6 cm spheres joined by a wire

Now, remember our puzzle? Here it is with numbers. Spheres of radii two centimetres and six centimetres, far apart, are joined by a thin wire and share eight microcoulombs; the Greek letter mu with C means microcoulomb. Find their charges, and the ratio of their surface fields. Pause the video and try it first. Done? Okay, let's check. At equal potentials, q is proportional to R; the curly sign means proportional to. So q one to q two is two to six. Right? Split eight microcoulombs in the ratio two to six, and the small sphere gets two microcoulombs, while the large sphere gets six microcoulombs. So the large sphere has three times the charge. With me? So, back to our puzzle. Three times the charge means the large sphere has the stronger field, right? No, no, wait, this is exactly the trap! The surface field goes as q over R squared, and q goes as R, so the field goes as one over R, okay? So, with the labels small and large below E, the ratio is six to two, which is three to one. The small sphere has three times the field, with one third of the charge! So our puzzle is solved. Clear?

6:21Example 2: potential of 8 merged drops at 10 V each

Now our second example. Eight identical drops, each at a potential of ten volts, merge into one. Find the potential of the big drop, and the ratio of its surface density to that of a small drop. Pause the video and try it first. Done? Okay, let's check. Eight times the charge, so eight times the potential, eighty volts, right? Oh no, careful, the radius grows too! With n equal to eight, V equals n to the power two thirds, times v. The cube root of eight is two, and two squared is four, so four times ten gives forty volts. Got it? And sigma prime is n to the power one third, times sigma, which is two sigma; the double arrow means so. So the surface densities are in the ratio two to one. Clear?

7:13Quick revision: field, cavity, spheres, pressure, drops

So, quick revision. Inside the metal, E is zero, net charge sits on the outer surface, and just outside, E is sigma over epsilon naught. Charge q in a cavity puts minus q on the wall and plus q on the outer surface. Joined spheres have q proportional to R, while sigma and E go as one over R, okay? The pressure is sigma squared over two epsilon naught, and merged drops reach a potential of n to the power two thirds, times v. Now go and try the practice problems on this page. You can do them. All the best!