Potential Energy and Work

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Four equal charges q sit at the corners of a square. To find their total potential energy, we add one term for every pair of charges. The square has four sides, so there are four pairs, right? Hmm, hold that thought. Potential energy, and the work needed to move charges around, come up often in JEE questions, so let's begin.

0:28Energy of a pair, and of a system

So, start with just two charges. Their potential energy, capital U, equals k q one q two over r. Here q one and q two, with the small subscripts one and two, are the charges, r is the distance between them, and k is one over four pi epsilon naught. And this energy is zero when the two charges are infinitely far apart. Okay? Now let me ask you something. Take a plus q and a minus q. Is their potential energy positive, or negative? It is negative, yes! Put q one and q two into the formula with their signs, and plus q times minus q is negative. Never drop the signs. Clear? Now, a system of many charges. Capital U equals the sum over pairs of k q i q j over r i j. The big Greek letter, capital sigma, means add up, and the word pairs below it means one term for every pair. Here q i and q j, with the small subscripts i and j, are any two of the charges, and r i j is the distance between them. Each pair is counted only once, so q one with q two, and q two with q one, is the same term. With me? And there is one more result here. If a charge q sits in an external potential V, made by other charges, its potential energy is simply q times V, right? Keep this one ready, because our second example needs it.

2:02Work and potential energy

Now, let's talk about work. Suppose you move a charge q slowly from point A to point B, so slowly that it gains no speed. The work done by you, the external agent, is written W ext, with the small word ext below it. It equals delta U, the change in potential energy; capital delta, the Greek triangle, means change. And that equals q times the bracket, V B minus V A, where the small B and A below each V name the two points, okay? Now let me ask you this. During the same move, the electric field is also pushing on the charge. So what is the work done by the field? It is minus delta U, yes, the same size with the opposite sign! So W field equals minus delta U, with that minus sign in front. And this is a common trap. Read carefully whether the question asks for the work by the field, or the work by an external agent, because the signs are opposite. Got it? One more unit you will meet. One e V, the electron volt, equals one point six times ten to the power minus nineteen joule; the capital J is the joule. It is the energy gained by a charge e, the size of an electron's charge, when it moves through one volt. Clear?

3:22Self-energy and energy in the field

Now, even one charged body stores energy, and we call it its self-energy. For a thin shell with charge capital Q and radius capital R, the self-energy is k capital Q squared over two capital R, okay? For a uniform solid sphere, it is three k capital Q squared over five capital R, a different fraction. Now, same charge and same radius. Which one has more self-energy, the shell or the solid sphere? The solid sphere, yes, because three fifths is more than one half. Finally, energy is stored in the electric field itself. The energy per unit volume, small u, equals one half epsilon naught E squared. Here E is the field, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. With me?

4:14Example 1: energy of four charges at the corners of a square

Now, remember our puzzle from the start? Let's finish it. Find the potential energy of four charges q at the corners of a square of side small a. Pause the video and try it first. Done? Okay, let's check. The square has four sides, so four pairs, each giving k q squared over small a. So the answer is four k q squared over small a, right? Oh no, careful, that is exactly the trap from our puzzle! Every charge pairs with every other charge, not just with its neighbours. So four charges give six pairs: the four sides, and the two diagonals. Okay? The four sides each have length small a, so together they give four times k q squared over small a. Right? The two diagonals are longer, root two times small a, with the square root sign on screen. So they give two times k q squared, over root two times small a. Clear? Now take k q squared over small a common. Inside the bracket we get four plus two over root two, and two over root two is just root two. So capital U equals k q squared over small a, times the bracket, four plus root two. Those two diagonals are what the puzzle missed. Got it?

5:35Example 2: work to bring a fifth charge to the square centre

Now a harder one, built on the same square. Four charges q sit at the corners of a square of side small a. How much work is needed to bring a fifth charge q from infinity to the centre of the square? Pause the video and try it first. Done? Okay, let's check. The work is q times the bracket, V centre minus V at infinity, with the small word centre below the V; the sideways eight on screen is infinity. The potential at infinity is zero, so the work is just q times V centre. This is our q times V from before. With me? Now we need the distance from the centre to each corner. Half the side, so small a over two, right? Oh no, careful, small a over two only takes you to the middle of a side! The corners are half the diagonal away, so root two times small a, over two, which is small a over root two. Clear? So each corner charge gives k q over the distance, small a over root two; the slash on screen is read as over. Potential is a scalar, so the four equal potentials simply add. Four times that gives four root two k q over small a. Okay? So the work is q times V centre, which is four root two k q squared over small a. It is positive, because we push the fifth charge in against the repulsion of four like charges. And notice, we never needed the energy of the square itself, only the potential at the centre. Got it?

7:09Quick revision: pairs, work signs, self-energy, field energy

So, quick revision. The energy of a system is the sum over every pair, each pair once, with signs. The work by an external agent is delta U, which is q times V B minus V A, and the work by the field is minus delta U, the opposite sign. Self-energy is k capital Q squared over two capital R for a shell, and three k capital Q squared over five capital R for a solid sphere, and the field stores one half epsilon naught E squared in each unit of volume. Now try the practice problems on this page. All the best!