Potentials of Continuous Distributions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Put a charge Q evenly on a full ring, of radius R, and look at its centre. Now put the same charge on half a ring, same radius, then on a quarter ring. Which one gives the biggest potential at the centre? Keep your answer in mind. Today we find the potentials of rings, rods, discs, shells and spheres; they come up often in JEE, so let's begin.

0:29One method: break the charge into tiny pieces

So, how do we handle charge that is spread out? Break it into tiny pieces. Each tiny charge, d q, at distance r, gives a tiny potential, d V, equal to k d q over r, where k is one over four pi epsilon naught. Epsilon naught, the epsilon with a small zero below it, is the permittivity of free space. Now, to add these, do we need components and directions, like for the field? We do not need them, because potential is a scalar, so we just add numbers. The long S sign is the integral: V equals the integral of k d q over r, and every result today comes from this line. With me?

1:12Ring and arc: every piece at the same distance

Look at part b of the figure: a ring of radius R, with centre O; ignore the field arrows and parts a and c. Every tiny charge d q is at the same distance R from the centre, so V centre equals k Q over R; the small word centre says where. Now, what if the charge is bunched up on one side, or sits only on an arc, like half a ring? Nothing changes at all, and that is the beauty of it! Every piece is still at distance R, and we only add numbers. So, back to our puzzle: the full ring, the half ring and the quarter ring all give k Q over R, exactly the same. Clear? Now take the point P on the axis, at distance x from the centre. The slanted line from d q to P is the square root of R squared plus x squared, by Pythagoras, the same for every piece. So V equals k Q over that square root; the tick-shaped sign is the square root, okay?

2:13Rod on its axis: a natural log appears

Next, a straight rod of length capital L, charged uniformly. Lambda, the Greek letter, is the linear charge density, the charge per unit length. Point P is on the axis, at distance d from the near end. Got it? Take a tiny piece of length d x, at distance x from P; its charge, d q, is lambda times d x. To cover the whole rod, from where to where must x run? Yes, exactly, from d at the near end to d plus capital L at the far end, and these are the limits on the integral sign. The integral of d x over x is the natural log, written l n, so V equals k lambda times the natural log of d plus capital L, over d. Right?

3:01Disc on its axis: add up rings

Now a disc of radius capital R, charged uniformly. Sigma, the Greek letter, is the surface charge density, the charge per unit area. Cut the disc into thin rings. One ring of radius small r and width d r has charge d q equal to sigma times two pi r d r; the dot means times. With me? Each ring gives k d q over the square root of r squared plus x squared. Add the rings from zero to capital R, and V is sigma over two epsilon naught, times the bracket: the square root of capital R squared plus x squared, minus x. Now a quick one: what is V at the centre, where x is zero? Very good if you got that one! Put x equal to zero, and the bracket becomes just capital R, so V equals sigma R over two epsilon naught. Got it?

3:52Shell and solid sphere

Now spheres. Look at part c of the figure, V against r; parts a and b belong to another lesson. First, take a thin shell of radius capital R, with charge Q, whose field E inside is zero. So, is the potential inside zero too? If you said yes, careful, because that is the trap! Zero field means the potential does not change, so inside, it stays constant at k Q over capital R, the flat orange line marked shell. Outside, V equals k Q over small r, the falling curve. Clear? Now a uniformly charged solid sphere. Inside it, V equals k Q times, in brackets, three capital R squared minus r squared, all over two capital R cubed. That is the dark curve marked solid, highest at the centre. Okay? Put small r equal to zero: V centre is three k Q over two capital R, and at the surface it is k Q over capital R, so the centre is three halves of the surface, one and a half times. The small words below V just say where. Got it?

5:05Infinite line of charge: differences only

Last, an infinitely long, straight line of charge, lambda per unit length. For points at distances r one and r two, the small one and two being labels, V one minus V two equals lambda over two pi epsilon naught, times the natural log of r two over r one. The catch: we cannot put the zero of potential at infinity here, so we find only differences. Right?

5:28Example 1: V and E at the centre of a non-uniform ring

Alright, example one. Charge Q is spread non-uniformly on a ring of radius R, so it is not even. Find V at the centre, and tell me whether the field E there is zero. Pause the video and try it first. Done? Okay, let's check. Every element is at distance R, and potentials add as scalars, so V equals k Q over R, however uneven the spread. Now for the field, which is a vector: for E to be zero, the tiny fields from all the pieces must cancel as vectors. With more charge on one side, do they cancel? They need not cancel, because the elements do not balance. So V is k Q over R, but E is not necessarily zero. Same potential, different field, so never mix the two up. Clear?

6:20Example 2: V on a rod's axis at distance L, 9 ln 2 volts

Now a slightly harder one. Here, a rod of length capital L has lambda equal to one nanocoulomb per metre, written n C, m to the power minus one. Find the potential on its axis, at distance capital L beyond one end. Pause the video and try it first. Done? Before we check, notice something strange: we are not told the value of capital L, okay? So can we still get a number for the potential? Yes, we can, because d, the distance from the near end, equals capital L, so x runs from capital L up to two times capital L, and the natural log of two L, over L, is just the natural log of two. So the length cancels. With me? Now the numbers. k is nine times ten to the power nine, and lambda is ten to the power minus nine coulomb per metre, so k lambda is just nine volts. Right? So V is nine times the natural log of two. The natural log of two is about zero point six nine three, and nine times that gives about six point two four volts; the wavy equals sign means approximately. Clear?

7:34Quick revision: ring, rod, disc, shell, sphere, line

So, quick revision. Ring or arc at the centre, k Q over R; rod on its axis, k lambda times the natural log of d plus L, over d. Disc centre, sigma R over two epsilon naught; shell, k Q over R everywhere inside. Solid sphere, centre one and a half times the surface; infinite line, differences only. Now go and try the practice problems on this page. All the best!