Energy Conservation for Moving Charges

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. Take two particles with the same charge, one of them twice as heavy as the other. Hold them close together, and let go. They push each other apart, and the stored energy turns into motion. So who gets that energy? All of it to the lighter one, or half and half? Hmm, hold that thought. Energy methods like this come up often in JEE problems on moving charges. So, let's begin.

0:30Energy conservation: K+qV stays the same

So, let us start with the main idea. When a charge moves in an electric field, its total energy stays the same. K i plus q V i equals K f plus q V f, okay? The small letters i and f below are labels, for initial and final. With me? Capital K is the kinetic energy, and q times capital V is the potential energy of the charge q, at a point where the potential is V, okay? So energy only changes form, from potential to kinetic, or back. Right? Now take a charge q that starts from rest and moves through a potential difference delta V; the capital delta, the Greek triangle, means a change. So where does its kinetic energy come from? Yes, exactly, it comes from the potential energy it loses! So for a charge of mass m, one half m v squared equals q times delta V, the charge times the potential difference. Remember this one line, it comes up often. Got it? Now let me ask you this. An electron is negative. If it starts from rest, does it move towards higher potential, or towards lower potential? It moves towards higher potential, yes! For a negative q, the potential energy q V falls as V rises, and what it loses becomes kinetic energy. So the electron speeds up while moving towards higher potential.

1:58Closest approach to a fixed nucleus

Clear? Next, a classic situation. Take a particle with charge q and kinetic energy capital K, moving head-on towards a nucleus of charge Z e. Capital Z is the atomic number, and e is the electronic charge. And the nucleus is fixed, it does not move. Okay? Far away, its kinetic energy K i is just K, and the potential of the nucleus there is zero. Right? Now, as it comes closer, the nucleus pushes it back, and it slows down. At the closest point, what is its speed? It is zero, exactly! Head-on, it stops for an instant at the closest distance r naught, the r with a small zero below, and then turns back. There the potential is k Z e over r naught, where k is one over four pi epsilon naught. With me? So energy conservation gives K equals k q Z e over r naught, and the double arrow means therefore. So r naught equals k q Z e over K, the kinetic energy. More energy, closer approach. But only for a head-on hit on a fixed nucleus, okay?

3:07Climbing a potential hill

Now picture a potential hill. Along the path, the potential rises from V start up to V top, and then falls again. The small words below are just labels. With me? The particle starts with kinetic energy K, and climbing to the top, it loses q times the rise in potential. So K top equals K, minus q times the bracket, V top minus V start. Right? Now tell me this. If the particle only just gets over the top, what is K top? It is zero, yes, and it can never be negative! So to pass, K must be greater than or equal to q times the bracket, V top minus V start; the sign with a line under it means greater than or equal to. With less than that, the particle turns back before the top. Clear?

3:58Two free particles

Let us look at one more case. So far, one charge was held fixed, so only the moving particle could gain kinetic energy. Okay? Now both charges are free to move, like the two particles in our puzzle. Is energy conservation alone enough to find their speeds? No, it is not enough, because one energy equation cannot give two unknown speeds. So we use conservation of momentum together with energy. And this is exactly where the trap in our puzzle hides. Got it?

4:29Example 1: speed of an electron accelerated through 100 V

Let's try our first example. Find the speed of an electron accelerated from rest through one hundred volts; the capital V after the number is the unit, volt. Pause the video and try it first. Done? Okay, let's check. Using e, the size of the charge, one half m v squared equals e times V, the charge times the potential difference. So v equals the square root of two times e times V, over m; that tick-shaped sign is the square root. Okay? Now look at this formula. Do we need the charge and the mass separately? No, we only need their ratio! For the electron, e over m is one point seven six times ten to the power eleven; the unit, C kg to the power minus one, is read coulomb per kilogram. Clear? Let's put the numbers in. v equals the square root of two, times one point seven six times ten to the power eleven, times one hundred. That is the square root of three point five two times ten to the power thirteen. With me? So v is approximately five point nine times ten to the power six metres per second; the wavy equals sign means approximately, and the unit, m s to the power minus one, is read metre per second. Just one hundred volts, and the electron moves at almost six million metres per second! Got it?

5:55Example 2: charges of mass m and 2m released from rest

Now, remember our puzzle? Here it is as a proper problem. Particles of masses m and two m, each with charge q, are held at a distance r and released. Find the final speed of the lighter one. Pause the video and try it first. Done? Okay, let's check. At the start, the potential energy is k q squared over r. So we give all of it to the lighter one: one half m v one squared equals k q squared over r, right? Oh no, careful, that is the trap! The heavier particle moves too. They start from rest, so the total momentum is zero, and it stays zero. In size, m v one equals two m v two, so v two equals v one over two; the small one and two below v are labels. Okay? Now let's write the energy equation. Far apart, the potential energy becomes zero, so all of k q squared over r turns into kinetic energy of both particles. That is one half m v one squared, plus one half times two m, times v two squared. Right? Put in v two equals v one over two. The second term becomes one quarter m v one squared, so the total is three quarters m v one squared. So the lighter one carries one half m v one squared, and the heavier one only one quarter m v one squared. Clear? So v one equals the square root of four k q squared, over three m r. So our puzzle is solved: the energy is not shared half and half, and the lighter one does not take it all. It moves twice as fast, and carries twice the kinetic energy of the heavier one, that is, two thirds of the total. Got it?

7:41Quick revision: K + qV, closest approach, hills, momentum

So, quick revision. K plus q V stays the same, and from rest, one half m v squared equals q times delta V, okay? Head-on at a fixed nucleus, r naught equals k q Z e over K, and to pass a hill, K must be at least q times the rise, V top minus V start. And when both particles are free, use momentum and energy together. Now try the practice problems on this page. All the best!