Capacitance and Standard Capacitors

Transcript

0:00Introduction

Hello everyone! Here is a big question. The Earth is a giant conducting ball, six thousand four hundred kilometres in radius. Treat the whole Earth as a capacitor. How big is its capacitance? Millions of farads? Hold that thought; we will work it out today. Capacitors come up often in JEE problems. So, let's begin.

0:25Capacitance: charge stored per volt

So, what is capacitance? Give a conductor a charge Q, and its potential becomes V; for two plates, V is the potential difference between them. The capacitance C equals Q over V, the charge stored per volt. Okay? Its unit is the farad, F for farad, and one farad is one coulomb per volt, written C V to the power minus one. Careful, this C is the coulomb, not the capacitance. Right? Now let me ask you this. If I double the charge on a capacitor, does its capacitance double too? The capacitance does not change at all, and this is a common trap. Double the charge, and V doubles too, so the ratio stays the same. C depends only on the geometry and on the medium in between, never on Q or V themselves. Clear?

1:14Isolated sphere and parallel plates

Let us start with the simplest one, an isolated sphere of radius R, okay? Its capacitance is four pi epsilon naught times the radius. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space, so this is in vacuum. With me? Now back to our puzzle. The Earth's radius is six point four times ten to the power six metre. So what is the capacitance of the Earth? Four pi epsilon naught is one over k, with k equal to nine times ten to the power nine, so C equals R over k. Six point four times ten to the power six, over nine times ten to the power nine, gives about seven hundred and eleven microfarad; the wavy sign means approximately, and mu means micro. Less than a thousandth of a farad, for the whole Earth, surprising, right? Now the most important one, the parallel plate capacitor of figure a: two plates, each of area A, a distance d apart. C equals epsilon naught A over d. Bigger plates give more capacitance; plates farther apart give less. Got it? Between the plates, the field E equals sigma over epsilon naught, where sigma, the Greek letter, is the charge per unit area on a plate. It also equals V over d, the potential difference over the gap. Clear?

2:39Spherical and cylindrical capacitors

Next, the spherical capacitor of figure b. The inner sphere, of radius a, carries plus Q, and the outer sphere, of radius b, is earthed, meaning connected to the ground. Then C equals four pi epsilon naught, times a times b, over b minus a. Okay? Now push the outer sphere very far away, so b tends to infinity; that is the arrow and the infinity sign. What does this formula become? Write it as four pi epsilon naught times a, times b over b minus a. As b grows, b over b minus a goes to one, so C tends to four pi epsilon naught times a, exactly the isolated sphere. Everything fits, right? But careful with one variation. If the inner sphere is earthed and the outer one is charged, C becomes four pi epsilon naught b squared, over b minus a. So always read which sphere is earthed. With me? And the cylindrical capacitor, of length L, with inner radius a and outer radius b. C equals two pi epsilon naught L, over the natural log of b over a; the l n on screen means natural log. Got it?

3:56The force between the plates

Now, the two plates carry opposite charges, so they attract. The field between them is E, and a plate carries charge Q, so the force is simply Q times E, right? No, no, wait, that is the trap! The force is only half of that: F equals Q squared over two epsilon naught A, which is one half of Q times E, okay? So why a half? Because each plate feels the field of the other plate only, never its own, and the other plate alone makes just half of the field. Clear?

4:30Two plates carrying Q₁ and Q₂

Now a situation that comes up often: two parallel plates whose charges are not equal and opposite. One carries Q one, the other Q two; the small one and two below are subscripts. So, how much charge sits on each outer face in figure c? Both outer faces carry the same charge, half of the total: Q one plus Q two, over two. Okay? And the inner faces, facing each other, carry equal and opposite charges, plus or minus Q one minus Q two, over two; that is the plus-minus sign. Right? So the charge facing across the gap is Q one minus Q two, over two, and dividing by C gives the potential difference: V equals Q one minus Q two, over two times C, got it?

5:18Example 1: C and Q for A = 0.02 m2, d = 1 mm at 100 V

Now, Example one. Parallel plates of area zero point zero two square metre, written m squared, are one millimetre apart. Find the capacitance, and the charge at one hundred volts. Pause the video and try it first. Done? Okay, let's check. One millimetre is ten to the power minus three metre. So C equals epsilon naught A over d, with epsilon naught equal to eight point eight five times ten to the power minus twelve. With me? Eight point eight five times zero point zero two is zero point one seven seven, so the top is zero point one seven seven times ten to the power minus twelve. Dividing by ten to the power minus three gives one point seven seven times ten to the power minus ten farad, or one hundred and seventy seven picofarad, p for pico. Right? Now the charge at one hundred volts. Which formula do we use here? Q equals C V, yes, exactly! One point seven seven times ten to the power minus ten, times one hundred, is one point seven seven times ten to the power minus eight coulomb. That is seventeen point seven nanocoulomb, n for nano. Easy, isn't it?

6:34Example 2: face charges and V for plates with 5 uC and -1 uC

Now a harder one. Two parallel plates, of capacitance one microfarad, carry plus five microcoulomb and minus one microcoulomb. Find the charges on all four faces, and the potential difference. Pause the video and try it first. Done? Okay, let's check. Q one is plus five and Q two is minus one; keep that minus sign! The outer faces get half the total: five plus minus one is four, and four over two gives two microcoulomb each. With me? Now the inner faces. Many students write five minus one here. Is that right, or is it five minus minus one? It is five minus minus one, which is six, so careful with the sign! Six over two is three, so the inner faces carry plus three and minus three microcoulomb. Right? So, with the plus five plate on the left, the four faces from left to right carry two, three, minus three and two microcoulomb. Together they add up to the total charge, four microcoulomb. Got it? Now the potential difference. One plate has five microcoulomb, and C is one microfarad, so V is five volts, right? That would be wrong, so careful! Only the charge facing across the gap counts, three microcoulomb. So V equals three microcoulomb over one microfarad, which is three volts. Clear?

8:03Quick revision: C formulas, face charges and plate force

So, quick revision. Capacitance is Q over V, fixed only by the geometry and the medium. Sphere, four pi epsilon naught R; plates, epsilon naught A over d; spherical, four pi epsilon naught, times a times b, over b minus a; cylinder, two pi epsilon naught L over the natural log of b over a. Two plates: outer faces carry half the total, inner faces plus or minus half the difference, and the force is one half of Q times E, okay? Now try the practice problems on this page. All the best!