The Electric Dipole

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. Hold a plus charge and an equal minus charge very close together. The total charge is zero. So far away, the electric field should be zero, right? Hmm, it is not quite that simple! This pair is an electric dipole, and it comes up often in JEE electrostatics. So, let's begin.

0:26What is an electric dipole?

So, what is a dipole? Two equal and opposite charges, plus q and minus q, written with the plus-minus sign, a distance two a apart. So the letter a is half the separation. Okay? We measure its strength by the dipole moment, a vector. So which way does it point? From plus q to minus q, or from minus q to plus q? It points from minus q to plus q, yes! Careful, do not draw it the other way. Its size is p equals q times two a; the brackets just group the two a. And p with the arrow on top is the vector itself, size and direction together. Right? And the unit is the coulomb metre, the C m on screen, a charge times a distance. Clear?

1:11Short dipole: axial and equatorial fields

For the field, k is one over four pi epsilon naught, where epsilon naught, epsilon with a small zero, is the permittivity of free space. Short dipole means r is much greater than a; that is the double greater-than sign. In the left picture, point A is on the axis and point B is on the bisector, both at distance r; the right one comes in the next video. With me? On the axis, the field is two k p over r cubed, along p, the same way as the dipole moment. Okay? Now the equatorial point, B, on the bisector. Which way does its field point? Along p, or opposite to p? Opposite to p, yes, exactly! On the bisector, the field is k p over r cubed, opposite to p. So at equal distances, the axial field is twice as strong, and the two fields point opposite ways. Students often get the size right and the direction wrong, so watch the direction. Got it?

2:09Any point: angle θ from the axis

Now any point, still far away, at distance r and angle theta from the axis; theta is a Greek letter. E r, with a small r below, is the part along the line from the centre: two k p cos theta over r cubed. E theta, with theta below, is at right angles to it: k p sine theta over r cubed. Right? Combine the two parts with Pythagoras. So E equals k p over r cubed, times the square root of one plus three cos squared theta. Clear? Now a quick check. Put theta equal to zero, which means the point is on the axis. What do you get? Cos zero is one, so one plus three gives four, and root four is two, giving two k p over r cubed, the axial field, perfect! And theta of ninety degrees gives k p over r cubed, the equatorial field. Okay? As for the direction, the field makes an angle alpha, another Greek letter, with r hat, the unit vector along r, shown by the little cap. And tan alpha is half of tan theta. Got it?

3:12Potential of a short dipole

Next, the potential of a short dipole, again far away: V equals k p cos theta over r squared, or in vector form, k times p dot r, over r cubed, with the dot product. Potential is a scalar, so no directions to add. With me? On the equatorial plane, theta is ninety degrees, and cos ninety degrees is zero; the double arrow means it follows that. So V is zero on that plane. Right? So let me ask you this. The potential is zero on the equatorial plane. Is the field zero there too? It is not, and this is a common trap. The field there is k p over r cubed; the slashed equals sign means not equal to. Zero potential does not mean zero field. Clear?

3:55How fast it dies away, and the exact fields

Look at the powers of r: the field falls as one over r cubed, the potential as one over r squared; the open alpha sign means proportional to. That is one power faster than a point charge. So, our puzzle: zero total charge, yet a field that just dies away faster. Okay? But all this holds only far away. Closer in, use the exact results. On the axis, E equals two k p r, over the bracket r squared minus a squared, whole squared. Got it? And on the bisector, E equals k p over the bracket r squared plus a squared, to the power three over two. Now, when r is much greater than a, what do these two become? Yes, exactly! Then a squared is tiny next to r squared, so we drop it, and get back two k p over r cubed and k p over r cubed.

4:44Example 1: dipole moment, axial and equatorial fields

Right? Now, Example one. Charges plus and minus two micro coulombs are two centimetres apart; the Greek letter mu means micro. Find the dipole moment, and the field twenty centimetres away, on the axis and on the bisector. Pause the video and try it first. Done? Okay, let's check. Two centimetres is zero point zero two metre, so p equals two times ten to the power minus six, times zero point zero two, which is four times ten to the power minus eight coulomb metre. Clear? With k as nine times ten to the power nine, k p is three hundred and sixty. Is it a short dipole? Yes, zero point two metre is far more than the half separation, zero point zero one metre. With me? Now r cubed is zero point zero zero eight. On the axis, two k p, which is seven hundred and twenty, over zero point zero zero eight, gives nine times ten to the power four newton per coulomb, written N C to the power minus one. Okay? And the equatorial field is half of that, four point five times ten to the power four newton per coulomb, opposite to p. So both fields are done, very good!

5:54Example 2: V and E at 60 degrees from a short dipole

Now a harder one. Find the potential V, and the modulus of E, shown by the two vertical bars, at r of ten centimetres and theta of sixty degrees, for a short dipole of moment ten to the power minus nine coulomb metre. Pause the video and try it first. Done? Okay, let's check. k p is nine times ten to the power nine, times ten to the power minus nine, which is nine. And r cubed is ten to the power minus three, so k p over r cubed is nine thousand. Right? Now the potential. Which power of r goes below here, r squared or r cubed? It is r squared, correct! The field has r cubed, the potential r squared. So V is k p cos theta over r squared, that is, nine times zero point five, over zero point zero one, which gives four hundred and fifty volts. Got it? Now the field: nine thousand, times the square root of one plus three cos squared sixty degrees. Cos sixty degrees is a half, so one plus three quarters gives one point seven five, and E is nine thousand times root one point seven five. Clear? The square root of one point seven five is about one point three two three. So E is about one point one nine times ten to the power four newton per coulomb, where the wavy equals sign means approximately. Well done if you got that one!

7:17Quick revision: dipole moment, fields and potential

So, quick revision. The dipole moment is q times two a, from minus q to plus q. The axial field of a short dipole is two k p over r cubed, along p; the equatorial field is k p over r cubed, opposite to p. The potential is k p cos theta over r squared: zero on the equatorial plane, though the field is not. Now try the practice problems on this page. All the best!