Mixed Problems: Set 2 (Potential, Energy and Dipoles)

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Inside a charged hollow shell, the electric field is zero. So the potential inside is also zero, right? Many students say yes, and lose marks. Hold on to that answer; we will test it soon. Today is our second set of mixed problems, on potential, energy and dipoles, and problems like these come up often in JEE Main. So, let's begin.

0:29Idea 1: potentials add as scalars

Let us start with idea number one. Charges of one, two and minus three microcoulomb sit at the corners of an equilateral triangle; mu C, with mu the Greek letter, means microcoulomb. Find the potential at the centroid. But the side is not given. Can we still do it? Yes, we can, and here is why. The potential of a point charge q at a distance r is k times q, over r, where k is nine times ten to the power nine in S I units. Okay? And potential is a scalar, so the potentials of several charges simply add, with their signs and no directions. With me? Now, the centroid of an equilateral triangle is at the same distance r from all three corners. Right? So V equals k over r, times one plus two minus three, times ten to the power minus six. The bracket is zero, so the potential is zero, whatever the side is, and that is the whole trick. Clear? Now let me ask you this. The potential here is zero. So the electric field there is zero too, right? No, no, careful, that is a trap! The field E, with the arrow on top because it is a vector, adds as a vector, and these unequal charges do not cancel. The slashed equals sign means not equal, so V equals zero does not mean E equals zero. Got it?

1:53Idea 2: inside a charged shell

Now, idea number two, our puzzle from the start. Here, a thin shell of radius ten centimetres carries five nanocoulomb, written n C on screen. Find the potential five centimetres from its centre, so inside the shell. The field inside is zero. So the potential is zero too, right? Let us check with the figure. Look at part c, the graph of V against r; parts a and b, the equipotentials, we skip today, and also the dark curve marked solid, which is for a solid sphere with charge spread evenly through its volume, not a metal one. Outside the shell, V is k Q over r, as marked on the right, with r the distance from the centre. And inside a thin shell, the field E is zero, yes. Okay? But no, the potential is not zero; that is the other half of the same trap. Zero field only means that V does not change as we move inside. So V stays at its value on the surface, k times capital Q, over capital R, where capital Q is the charge on the shell and capital R is its radius. That is the flat orange line marked shell. Clear? Now the numbers, in S I units. Nine times ten to the power nine, times five times ten to the power minus nine, gives forty five. Divide by zero point one metre, and we get four hundred and fifty volts. Right? And do not put in the distance, zero point zero five metre; that gives nine hundred volts, which is wrong. Inside the shell, always use the radius. So E equals zero does not mean V equals zero either, and that solves our puzzle from the start. Got it?

3:36Idea 3: torque on a dipole

Next, idea number three: dipoles. Look at part b of the figure: plus q and minus q in a uniform field E, at an angle theta, the Greek letter theta, to it; part a we skip today. Here, a dipole of four times ten to the power minus nine coulomb metre is at sixty degrees to a field of two times ten to the power four newton per coulomb. Find the torque. So, do we use the sine, or the cosine? The sine, that is correct! The forces q E and minus q E form a couple, with no net force, and the torque tau, the Greek letter like a t, equals p times E times sine theta, where p is the dipole moment. The cosine belongs to the energy, minus p E cos theta, written under the figure, so do not mix them up. Okay? Now p times E is four times ten to the power minus nine, times two times ten to the power four, which gives eight times ten to the power minus five. Right? Sine sixty degrees is root three over two. So tau is eight times ten to the power minus five, times root three over two, about six point nine times ten to the power minus five newton metre; the wavy equals sign means approximately. Got it?

4:55Example 1: potential energy of 2 uC and -3 uC 30 cm apart

Alright, example one. Find the potential energy of two microcoulomb and minus three microcoulomb, held thirty centimetres apart. Pause the video and try it first. Done? Okay, let's check. With Coulomb's force, we often put in only the sizes of the charges. So here also we drop the minus sign, right? No, no, wait, energy is a scalar, and its sign carries the meaning. So U equals k times q one times q two, over r, with the signs kept; q one and q two, with the small one and two below, are the two charges. And r must be in metres: thirty centimetres is zero point three metre. Okay? Two microcoulomb times minus three microcoulomb gives minus six times ten to the power minus twelve. So U equals nine times ten to the power nine, times minus six times ten to the power minus twelve, over zero point three. Right? Nine times minus six is minus fifty four, so the top becomes minus fifty four times ten to the power minus three. Divide that by zero point three, and we get minus zero point one eight joule. Keep the minus sign: the energy is zero when the charges are infinitely far apart, and here it is below that, because the charges attract. Very good, well done!

6:17Example 2: E from V = 4x squared - 3y at (1, 1), root 73 V/m

Now a harder one, JEE level. The potential is V equals four x squared minus three y, in volts. Find the modulus of E, the size of the field, at the point one metre, one metre; the two straight bars mean modulus. Pause the video and try it first. Done? Okay, let's check. Many students take only the derivative with respect to x. Minus the derivative of four x squared is minus eight x, so at x equals one the field is minus eight, and its size is eight, right? Oh no, careful, because V depends on y as well, so the field has a y part too. E x, the E with a small x below, is minus the partial derivative of V with respect to x, and E y is minus the partial derivative with respect to y. The curly d means a partial derivative: differentiate in one variable, holding the other fixed. With me? So E x is minus eight x. And V has minus three y, so its partial derivative with respect to y is minus three, and E y is minus of minus three, which is plus three. Clear? At the point one, one, the field has x part minus eight and y part three, in volts per metre, written V m to the power minus one on screen. Okay? The size is the square root of, eight squared plus three squared. Sixty four plus nine is seventy three, so the answer is root seventy three, about eight point five volts per metre, and not eight. You see how the y part changed the answer?

7:55Quick revision: shell V, pair energy, torque, E from V

So, quick revision. Potentials add as scalars, and inside a shell, V is k Q over capital R, not zero. The energy of a pair is k q one q two over r, with the signs kept, and the torque on a dipole is p E sine theta. And for the field from V, take minus the partial derivative along each axis, then combine the parts. Now try the practice problem on this page. All the best!