A Dipole in an External Field

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Put an electric dipole, a plus charge and a minus charge close together, in a uniform electric field. The field pushes the plus charge one way, and the minus charge the other way, with forces of the same size. So the net force on the dipole is zero. Then does the dipole just sit there, doing nothing? Hmm, hold that thought. Dipoles in a field come up often in JEE problems. So, let's begin.

0:32Uniform field: zero net force, but a torque

So, here is our dipole in a uniform field E; the little arrow on top means it is a vector. The plus q feels a force q E along the field, and the minus q feels minus q E, against the field. Equal in size, opposite in direction, so the net force is zero. Okay? Now the real question from our puzzle. The net force is zero, but can the dipole still turn? Yes, it can, and it does! The two forces act at different points, so they form a couple, which gives a torque. Tau vector equals p vector cross E vector. Tau, the Greek letter, is the torque, p is the dipole moment, and the cross sign means the cross product. Clear? In size, tau equals p E sine theta, where theta is the angle between p and E, okay? So no net force, but it turns, and that answers our puzzle! Now tell me, at which angles does this torque become zero? The torque is zero when sine theta is zero, exactly! That is at theta equals zero, and at theta equals one hundred and eighty degrees; the small circle up there is the degree sign. So the dipole lies along the field, or exactly against it.

1:45Potential energy and work

Got it? Now let us look at the energy. The potential energy U equals minus p vector dot E vector; the dot means the dot product. That is minus p E cos theta. Remember that minus sign, okay? Let me ask you this. At which angle is this energy the lowest? At theta equals zero, yes! Cos zero is one, so U equals minus p E, the minimum. So the dipole lined up with the field is stable. Right? And at one hundred and eighty degrees, cos is minus one, so U equals plus p E, the maximum. This one is unstable: a tiny push, and the dipole flips over. Clear? Now let us talk about work. To turn the dipole from theta one to theta two, the work needed is p E times the bracket, cos theta one minus cos theta two; the small one and two below theta are just labels. It is U final minus U initial. With me? Quick question for you. How much work turns it from zero all the way to one hundred and eighty degrees? Cos zero is one, and cos one hundred and eighty is minus one. One minus minus one is two, so W equals two times p E, and this comes up often. Okay? Here is one more result. Turn the dipole slightly from theta equals zero and let it go, and it oscillates. Its angular frequency, omega, equals the square root of p E over capital I, the moment of inertia; the tick-shaped sign is the square root. Only for small angles, right?

3:19Non-uniform field: a net force

Now, what if the field is not uniform? Is the net force still zero? No, it is not! The two charges now sit in fields of different strength, so the forces do not cancel. For a dipole along the field, F equals p times d E over d x, how fast the field changes with distance. Okay? And the direction is easy to remember: a dipole lined up with the field is pulled towards the stronger field. Right? Now put a point charge q on the axis of a short dipole, at a distance r much bigger than the length of the dipole. The force is two k p q over r cubed, where k is one over four pi epsilon naught. Clear? Now be careful with this one. Does this force fall as one over r squared, like Coulomb's law? No, it does not, and this is a common trap! It falls as one over r cubed, one power faster. And for two short dipoles on the same axis, with moments p one and p two, it is faster again. F equals six k p one p two over r to the power four; the small one and two below p are labels. With me?

4:29Example 1: torque and energy of a dipole at 30 degrees

Let's try our first example. We have a dipole of ten to the power minus eight coulomb metre, that is the C m on screen, making thirty degrees with a field of ten to the power four newton per coulomb, the N C to the power minus one. Find the torque and the energy. Pause the video and try it first. Done? Okay, let's check. First, p times E: ten to the power minus eight, times ten to the power four, gives ten to the power minus four; the cross sign here simply means times. We need it in both answers. Right? The torque uses sine. Tau equals p E sine thirty degrees, so ten to the power minus four times one half. That gives five times ten to the power minus five newton metre, the N m on screen. Clear? Now the energy. Before calculating, tell me: is it positive or negative? It is negative, yes! The angle is less than ninety degrees, so cos is positive, and the minus sign stays. U equals minus p E cos thirty degrees, that is minus ten to the power minus four, times zero point eight six six. Okay? So U is approximately minus eight point seven times ten to the power minus five joule; the wavy equals sign means approximately, and J is joule. Torque with sine, energy with cos, and never drop that minus sign. Got it?

5:53Example 2: work to turn a dipole from 60 to 180 degrees

Now a slightly harder one. How much work turns a dipole from sixty degrees to one hundred and eighty degrees with the field? The dipole moment p is two times ten to the power minus six coulomb metre, and the field E is ten to the power five newton per coulomb. Pause the video and try it first. Done? Okay, let's check. Here theta one is sixty degrees and theta two is one hundred and eighty degrees. So W equals p E times the bracket, cos sixty minus cos one hundred and eighty. With me? Now p times E: two times ten to the power minus six, times ten to the power five, gives zero point two joule. Got it? Now the bracket. Cos sixty is one half, and then we take away one, so the bracket is minus one half, right? Oh no, careful, that is the trap! Cos one hundred and eighty is minus one, not one. So the bracket is one half minus minus one, that is one half plus one, which gives one point five. Clear? So W equals zero point two times one point five, which is zero point three joule. It is positive, because we turn the dipole away from its stable position. Okay? Here is a quick check, using energy. At sixty degrees, U is minus zero point one joule, and at one hundred and eighty degrees, it is plus zero point two joule. The rise is zero point three joule, the same answer, perfect!

7:18Quick revision: torque, energy, work, non-uniform field

So, quick revision. In a uniform field, the net force is zero, but the torque is p E sine theta. The energy is minus p E cos theta, and the work is p E times the bracket, cos theta one minus cos theta two. And in a non-uniform field, a dipole lined up with the field is pulled towards the stronger field. Now try the practice problems on this page. All the best!