Equilibrium of Charges

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two positive charges, plus four microcoulomb and plus one microcoulomb, sit thirty centimetres apart. Where on the line joining them can a third charge feel no force? Right in the middle, at fifteen centimetres, since that is fair to both, right? Hold that answer for a few minutes. Balancing charges comes up often in JEE, so let's begin.

0:28Like charges: where is the balance point?

So, here is the setup. Two like charges, meaning charges of the same sign, q one and q two, with the small subscripts one and two. They are point charges in vacuum, a distance d apart. Okay? Now we bring in a third charge, small q, on the line joining them. Let x be its distance from q one. First question for you. Can the balance point lie outside, beyond one of the two charges? It cannot, because outside, both forces act in the same direction, so they never cancel. Between the charges the forces are opposite, so set their sizes equal: k times q one times q, over x squared, equals k times q two times q, over, d minus x, the whole squared. Here k is one over four pi epsilon naught, with epsilon naught the permittivity of free space. With me? Now a topper's trick. Do not expand this into a quadratic. k and q cancel, and both sides are perfect squares, so take the square root, the root sign you see on screen. x over, d minus x, equals root q one over root q two, with no quadratic at all. Clear? Cross multiply and rearrange. x equals root q one over, root q one plus root q two, times d, from q one. If q one is the bigger charge, x is more than half of d, so the point sits nearer the smaller charge. Right? Now tell me one more thing. Does the sign of the third charge matter here? It does not matter at all, and the reason is on the screen: small q cancels from both sides. If it is negative, both forces become pulls, but they are still opposite and equal. So a third charge of any sign is in equilibrium there. Got it?

2:20Unlike charges: where is the balance point?

Now let us flip one sign. Two unlike charges, one positive and one negative. Can the balance point still lie between them? It cannot, and here is the reason. Take a positive charge between them. The positive charge pushes it towards the negative one, and the negative charge pulls it the same way. Both forces point the same way, so there is no balance. Okay? So the point must be outside. But on which side? Beyond the bigger charge, or beyond the smaller one? It lies beyond the smaller charge, yes, exactly! Out there, the smaller charge is nearer and the bigger one is farther, so their opposite forces can match. Beyond the bigger charge, the bigger one is also nearer, so its force always wins. With me? And the distance from the smaller charge is x equals root q small over, root q big minus root q small, times d. The words small and big are written as subscripts on q, okay? And one warning. Put in only the sizes of the charges, not their signs, since we are matching the sizes of two opposite forces. Clear?

3:31The whole system in equilibrium

Now a harder demand. The like charges q one and q two are also free, and we want all three in equilibrium. So now, what sign must the third charge have? It must be opposite to q one and q two, exactly! Look at q one. Charge q two already pushes it away. If the third charge also pushed, q one would fly off. So the third charge must pull it back, and that needs the opposite sign. Right? Balancing the forces on q one then gives the size of the third charge. q equals minus q one times q two, over, root q one plus root q two, the whole squared, at the same point as before. In example two, we will get the same answer by balancing the forces directly. Okay?

4:18Stable or unstable?

Now, equilibrium is not the whole story. Two like charges are fixed, and a third charge of the same sign rests on the line between them. We nudge it a little along the line, towards one charge. What happens? The nearer charge now pushes harder, so it is pushed back towards the balance point. So it is stable along the line, yes! But nudge it sideways, across the line, and both pushes have outward parts, so it moves away. So it is unstable across the line. Got it? And for a third charge of the opposite sign, everything flips. Along the line, the nearer charge pulls harder, so it runs away. Across the line, the two pulls bring it back. So never assume an equilibrium is stable; check both directions. With me?

5:06Example 1: null point of 4 micro C and 1 micro C 30 cm apart

Now, remember our puzzle from the start? Let's settle it. Charges plus four microcoulomb and plus one microcoulomb are thirty centimetres apart. The mu C on screen is microcoulomb, with the Greek letter mu meaning micro. Where should a third charge be placed so that it feels no force? Pause the video, and try it first. Done? Okay, let's check. Both are like charges, so the point is between them. Let x be its distance from the four microcoulomb charge, in centimetres. Balance the forces, cancel k and the third charge, and four over x squared equals one over, thirty minus x, the whole squared. Okay? Now the trick: take square roots. That gives two times, thirty minus x, equals x, and the double arrow means this leads to x equals twenty centimetres. Right? So the answer is twenty centimetres from the four microcoulomb charge, only ten from the smaller one. So our midpoint guess from the puzzle was a trap! At the midpoint, the bigger charge pushes four times harder. The point sits nearer the smaller charge, just as the formula said. Clear?

6:22Example 2: charge q that keeps 4e, e and q in equilibrium

Now a harder one. Charges four e and e are placed at a distance d, where e stands for a charge, so four e is four times it. What charge q, placed where, keeps all three in equilibrium? Use what we just learnt. Pause the video, and try it first. Done? Okay, let's check. Both are like charges, so the third charge goes between them, at our usual point. The root of four e is two root e, so x equals two over, two plus one, times d, which is two d over three from the four e charge. With me? So the third charge is d over three away from the charge e, as the second line shows. Now, the third charge is already balanced at this point. So to find its size, which charge should we balance next? We balance one of the outer charges, so let us take the charge e, the smaller one. The push from four e is k times four e times e, over d squared. The force from q is k times q times e, over, d over three, the whole squared. Taking the outward direction as positive, for balance their sum is zero. Okay? Cancel k, e and d squared, and four e plus nine q equals zero. So q equals minus four e over nine, at d over three from the charge e, between the two. The minus sign means it is opposite to both, exactly as we reasoned. Clear?

7:54Quick revision: null point and the whole system balance

So, quick revision. Like charges: the balance point lies between them, nearer the smaller charge. Unlike charges: it lies outside, beyond the smaller charge. And to balance the whole system, the third charge has the opposite sign, minus q one times q two, over, root q one plus root q two, the whole squared. Now try the practice problems on this page. All the best!