Fields of Continuous Charge Distributions

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Take a ring of charge, and walk along its axis from the centre. Where is the field the strongest? Most students say at the centre, where the charge is closest. Is that right? Hmm, hold that thought, because by the end you will find the exact spot. Rings and rods come up often in JEE, so let's begin.

0:25Spread-out charge: cut it into point charges

So far, we studied point charges, but charge is often spread out, along a rod or around a ring. For a line of charge, lambda, a Greek letter, is the charge per unit length; for a uniform rod, it is Q over L, okay? Now the main idea. Cut the charge into tiny pieces d q; each is a point charge, so d E equals k times d q over r squared. Here k is one over four pi epsilon naught, Greek epsilon with a small zero below, the permittivity of free space, in vacuum. With me? Each little field has its own direction, so they add as vectors: E vector is the integral of d E vector. The long S sign is the integral, and the arrows mean vectors. And first, use symmetry to cancel parts in pairs. Got it?

1:17A rod, and an infinite line of charge

Let's start with a straight rod, part a of the figure. Point P is at perpendicular distance r from it, with N the foot of the perpendicular, and the ends are seen at angles theta one and theta two from the perpendicular P N, Greek theta with small subscripts. Okay? The part of the field perpendicular to the rod, E with a small upside down T below it, is k lambda over r, times sine theta one plus sine theta two. For a positive rod, it points away. Clear? The part along the rod, E with two small parallel lines below, is k lambda over r, times cos theta two minus cos theta one. Right? Now stretch the rod to infinity, both ways. What happens to theta one and theta two? Both become ninety degrees, and the small circle means degrees. So the two sines add up to two, both cosines are zero, and the part along the rod vanishes. With me? So an infinite line gives E equals two k lambda over r, or lambda over two pi epsilon naught r. Careful, it falls as one over r, not one over r squared; the curly sign means proportional to. Okay?

2:29A ring, on its axis

Now a ring of radius R and charge Q, part b of the figure. Point P is on the axis, and x is its distance from the centre O, as marked. The piece d q at the top gives d E vector at P, slanting a little downward. Right? Now take the piece exactly opposite, at the bottom of the ring. What does its field at P do? It slants upward by the same amount! So the sideways parts cancel in pairs, and only the axial parts, d E cos phi, survive. Here phi, the Greek letter, is the angle between d E vector and the axis. Clear? Adding them, E equals k Q x, over x squared plus R squared, to the power three over two, along the axis, away from the ring for positive charge. Now our puzzle: is this largest at the centre, where the charge is closest? Oh no, careful, just put x equals zero: the top becomes zero, so E is zero at the centre, since opposite pieces cancel. Graph c shows the story: zero at the centre, a peak at the dashed line, R over root two, then a fall. We prove it in Example two.

3:44Arcs, discs and sheets

Next, the field at the centre of an arc of radius R that makes an angle two phi there, so here phi is half that angle. The field is two k lambda sine phi over R, along the bisector, the line cutting the arc into equal halves. Okay? For a semicircle, phi is ninety degrees and sine phi is one, so E is two k lambda over R, or lambda over two pi epsilon naught R, as shown. For a full ring, the angle at the centre is three hundred and sixty degrees, so phi is one hundred and eighty degrees, sine phi is zero, and the field is zero. Right? For a disc, Greek sigma is the charge per unit area. On its axis, E is sigma over two epsilon naught, times one minus x over root of x squared plus R squared. Let R go to infinity, shown by the arrow, and the fraction vanishes, so an infinite sheet gives sigma over two epsilon naught. With me?

4:44Example 1: axial field 20 cm beyond a 4 micro C, 20 cm rod

Now our first example. Here, a rod twenty centimetres long carries four microcoulomb uniformly. Find the field on its axis, twenty centimetres beyond one end. The Greek letter mu means micro, ten to the power minus six. Pause the video and try it first. Done? Okay, let's check. Can we treat the whole rod as a point charge at its middle? No, careful, we cannot! The near end is twenty centimetres away and the far end forty, so each piece has its own distance. Measure x from our point: a piece d x carries lambda d x, and gives k lambda d x over x squared. All point one way, so integrate from d to d plus L, to get k lambda times one over d, minus one over d plus L, okay? Since lambda times L is the total charge, Q, this becomes k Q over d times d plus L, as you see. Now use metres: d is zero point two metre, and d plus L is zero point four metre. Right? So E equals nine times ten to the power nine, times four times ten to the power minus six, over zero point two times zero point four. That gives four point five times ten to the power five newton per coulomb, away from the rod, as the charge is positive. Got it?

6:08Example 2: where the axial field of a ring is maximum

Now, our puzzle as a proper problem. Find where the axial field of a ring, of radius R and charge Q, is largest, and its value. Here is a hint: use calculus. Pause the video and try it first. Done? Okay, let's check. Since k Q is constant, we differentiate x over x squared plus R squared, to the power three over two; d over d x means the derivative with respect to x. This gives R squared minus two x squared, over x squared plus R squared, to the power five over two. Okay? For the largest field, this derivative must be zero. So which part of the fraction must be zero? Yes, the top, since the bottom is never zero. So two x squared equals R squared, and x is R over root two; the double arrow means this gives. And graph c confirms this turning point is the peak. Clear? Now the value. At x equals R over root two, x squared plus R squared is three R squared over two, so the bottom becomes three over two, to the power three over two, times R cubed. With me? Put it together: E max, with the small word max below, equals two k Q over three root three R squared, or Q over six root three pi epsilon naught R squared. So the field is largest not at the centre, but at R over root two. That solves our puzzle!

7:37Quick revision: line, semicircle, sheet and ring formulas

So, quick revision. Cut into tiny pieces, add the fields as vectors, use symmetry. An infinite line gives two k lambda over r, a semicircle two k lambda over R, an infinite sheet sigma over two epsilon naught. And the ring formula gives zero at the centre, largest at R over root two. Now try the practice problems on this page. All the best!