Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Four charges sit around the origin, all of the same size, all at the same distance: two positive on the positive axes, and two negative on the negative axes. So, by symmetry, the field at the origin must be zero, right? Hold that answer; our last example settles it. This is part A of our electrostatics problem bank, at JEE Main level: three quick tools, then two full problems. So, let's begin.

0:31Tool 1: flux through a closed surface

So, tool one, on Gauss's law. Here, a closed surface contains two microcoulombs and minus three microcoulombs; mu, the Greek letter, means micro, and capital C is the coulomb. Then a five microcoulomb charge sits outside. Find the flux, okay? For a closed surface, Gauss's law says phi, the flux, equals q in over epsilon naught. Here q in, the q with the little word in below it, is the charge inside; epsilon naught, epsilon with a small zero below, is the permittivity of free space. Clear? Now tell me, the five microcoulombs sitting outside, do they count as part of q in? No, they do not, that is correct! Only the charges inside count, so q in is two plus minus three, which gives minus one microcoulomb, or minus ten to the power minus six coulomb. With me? So phi is minus ten to the power minus six, over eight point eight five times ten to the power minus twelve. That is about minus one point one three times ten to the power five, newton metre squared per coulomb; the wavy sign means approximately. Right? But the outside charge still matters. It changes E vector, the E with an arrow on top, the field on the surface. It just does not change phi, the total flux. Got it?

1:50Tool 2: a charged conducting sphere

Now tool two. Here, a conducting sphere of radius ten centimetres has sigma equal to eight point eight five microcoulombs per metre squared; sigma, a Greek letter, is the charge per unit area. Find the field at its surface and its potential, alright? Just outside a conductor, E equals sigma over epsilon naught: eight point eight five times ten to the power minus six, over eight point eight five times ten to the power minus twelve. Right? The eight point eight fives cancel, so E is ten to the power six newtons per coulomb, written N C to the power minus one. Nice, isn't it? For the potential, V equals k capital Q over capital R, and E equals k capital Q over capital R squared, where k is one over four pi epsilon naught. Divide them, and V equals E times capital R, clear? So capital R is ten, and V is ten to the power six times ten, which gives ten to the power seven volts, right? No, no, wait, that is the trap! The radius must be in metres, and ten centimetres is zero point one metre. With me? So V is ten to the power six times zero point one, which gives ten to the power five volts; the upright V after the number is the volt. Got it?

3:01Tool 3: capacitors in series and parallel

Now tool three, on capacitors. Two microfarads and four microfarads in series are in parallel with three microfarads, across twelve volts; mu F is the microfarad, the unit of capacitance. Find the charges, okay? First, the series pair. One over capital C is one over two plus one over four, which gives three quarters, so capital C is four thirds of a microfarad. Clear? Now think about this: is the charge on this pair shared between the two capacitors, half and half? No, it is not shared, and that is the trap! In series, the same charge Q sits on each capacitor. So Q equals four thirds times twelve, which gives sixteen microcoulombs on each. Right? Let us check: sixteen over two gives eight volts, sixteen over four gives four volts, and eight plus four is twelve, the full battery voltage. In parallel, the voltage is the same, so the three microfarad capacitor gets the full twelve volts, and three times twelve gives thirty six microcoulombs. So the charges are sixteen, sixteen and thirty six microcoulombs. With me?

4:07Example 1: work done by field 3i + 4j on a 2 C charge

Now, example one. In a uniform field, E vector equals three i hat plus four j hat, newtons per coulomb; i hat and j hat, the letters with little caps, are unit vectors along x and y. Then a two coulomb charge moves from the origin to the point two metres, one metre. Find the work done by the field. Pause the video and try it first. Done? Okay, let's check. In a uniform field, the force, q times E vector, is constant, so W equals q times E vector dot delta r vector. Delta, the little triangle, means change, so delta r vector is the displacement. Clear? Now, a shortcut many students try. The size of E vector, the straight bars, is five, and the size of delta r vector is root five. So W is two times five root five, right? No, no, careful, that is the trap! That works only when the field and the displacement point the same way; here they do not. So, from the origin to two metres, one metre, delta r vector is two i hat plus j hat, in metres. Okay? For the dot product, multiply the matching parts and add: three times two is six, and four times one is four, so the dot product is ten. Right? So W equals two times ten, which gives twenty joules; capital J is the joule. Not two times five root five, because direction matters. Got it?

5:35Example 2: field at origin due to charges q, q, -q, -q

And now, our puzzle from the start, in the bank's words. Charge q sits at small a, zero and at zero, small a, on the positive axes. And minus q sits at minus small a, zero and at zero, minus small a, on the negative axes. Find the field at the origin. Pause the video and try it first. Done? Okay, let's check. Every charge is at distance small a from the origin, so each alone gives a field of size k q over small a squared, with k equal to one over four pi epsilon naught. Okay? Now take the pair on the x axis: plus q on the right, minus q on the left. Do their fields at the origin cancel? No, they add, and this is where our puzzle breaks! Plus q on the right pushes away, so its field at the origin points along minus x. Minus q on the left pulls towards itself, so its field also points along minus x. Clear? So the x pair gives two k q over small a squared, along minus x. In the same way, the y pair gives two k q over small a squared, along minus y. With me? These two equal fields are at right angles, so the resultant is root two times one of them: E equals two root two k q over small a squared. Right? And it points along minus of, i hat plus j hat, midway between minus x and minus y. So, for this arrangement, the field is not zero at all; the symmetry fooled us. Draw each direction first, then add.

7:12Quick revision: Gauss law, V = ER, capacitors, work done

So, quick revision. Gauss's law: phi equals q in over epsilon naught, inside charges only. For a conducting sphere, E equals sigma over epsilon naught, and V equals E times capital R, in metres. In series, the same Q; in parallel, the same V; in a uniform field, W equals q times E vector dot delta r vector. For many charges, draw each direction, then add as vectors. Now try the practice problem on this page. All the best!