Energy Stored and Charge Sharing

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You connect an empty capacitor to a battery. The battery pushes charge Q through a potential difference V, so the work it does is Q times V, the charge times the voltage. How much of that work ends up stored in the capacitor? All of it, surely? Hold that thought; the answer is surprising. Energy and charge sharing come up often in JEE questions. So, let's begin.

0:31Energy stored in a charged capacitor

So, the main result first. Capital U is the energy stored, capital Q the charge, capital C the capacitance, and capital V the potential difference. U equals Q squared over two C, or one half C V squared, or one half Q V, all equal because Q equals C times V, so use whichever form fits what you know. Okay? Next, energy density, written as a small u: the energy stored per unit volume. It equals one half epsilon naught E squared, where epsilon naught is the permittivity of free space, and capital E is the electric field, in vacuum or air. Got it?

1:10Charging from a battery: where does half go?

Now, back to our puzzle. Say a battery of voltage V charges a capacitor. It moves a total charge Q, equal to C times V, through the potential difference V, so the work W it does equals Q times V, with me? So, how much of this work gets stored in the capacitor? Only half, one half Q V, which is one half C V squared! The other half is lost as heat: Q V minus one half Q V gives one half C V squared of heat. Got it? Here comes a trap. Suppose you use wires with a much smaller resistance. Less resistance means less heat, right? No, and that is exactly where students slip: the heat is one half C V squared, whatever the resistance R is. Small resistance or big, half the battery's work is always lost. That solves our puzzle from the start.

2:09Charge sharing: joining two capacitors

Clear? Now, charge sharing. Take a capacitor C one at potential V one, and another, C two, at V two; the small one and two below are just labels. We join positive plate to positive plate, and negative to negative: like plates. Charge flows until both reach one common potential, and no charge escapes, so charge is conserved. Okay? So tell me, what is the total charge on the two capacitors? Yes, it is C one V one, plus C two V two! After joining, it sits on C one plus C two. So the common potential V equals C one V one, plus C two V two, all over C one plus C two. Clear? Now, charge is conserved. Is the energy conserved too? No, energy is lost, just as in the battery case! The energy lost equals C one C two, times the bracket V one minus V two, squared, all over two times the bracket C one plus C two. It is zero only when V one equals V two. Right? And if the positive plate of one is joined to the negative plate of the other, that is joining unlike plates. Then simply replace V two by minus V two in both formulas. Got it?

3:29Pulling the plates apart

Now the last idea: pulling apart the plates of a parallel plate capacitor. Remember, C equals epsilon naught A over d, where capital A is the plate area and small d is the gap. So a bigger gap means a smaller capacitance. Okay? First case: the battery is disconnected, so the capacitor is isolated. We pull the plates apart. Does the stored energy go up or down? It goes up, and here is why. With the battery gone, Q is fixed. Then U equals Q squared over two C, and C falls, so U rises. The symbol that looks like an open alpha means proportional to: U is proportional to d. The extra energy comes from the work done in pulling the plates apart. With me? Second case: the battery stays connected, so V is fixed. Now U equals one half C V squared, and C falls, so U falls: U is proportional to one over d. Same action, opposite answer! So always check the battery first. Right?

4:35Example 1: common potential and energy lost, 2 and 3 uF

Now one from the course. Here is the question: a two microfarad capacitor, charged to one hundred volts, is joined to an uncharged three microfarad capacitor. Find the common potential, and the energy lost. On screen, mu, the Greek letter, means micro, so mu F is microfarad. Pause the video and try it first. Done? Okay, let's check. The charge on the first capacitor is two microfarad times one hundred volts, which is two hundred microcoulomb, and the uncharged one has none. Right? After joining, these two hundred microcoulomb sit on two plus three, that is five microfarad. Two hundred over five gives forty volts. Clear? Here V two is zero, so the loss is C one C two times V one squared, over two times C one plus C two. And C one C two over C one plus C two is two times three over five, one point two microfarad. With me? So the loss is one half times one point two times ten to the power minus six, times ten to the power four, which is one hundred squared. That gives six times ten to the power minus three joules, or six millijoules. Got it? Let us double check with the energies. Before joining, one half times two microfarad times one hundred squared gives ten millijoules. After, one half times five microfarad times forty squared gives four millijoules. Ten minus four is six, a perfect match. Okay?

6:10Example 2: unlike plates joined, 100 V and 200 V

Now a harder one. Here is the question: a two microfarad capacitor at one hundred volts, and a three microfarad capacitor at two hundred volts, are joined positive plate to negative plate. Find the common potential and the energy lost. Pause the video and try it first. Done? Okay, let's check. Two times one hundred gives two hundred microcoulomb, and three times two hundred gives six hundred microcoulomb. With me? So the total charge is two hundred plus six hundred, which is eight hundred microcoulomb, right? No, and this is exactly the trap, because positive is joined to negative, the charges on the joined plates partly cancel. The net charge is six hundred minus two hundred, which is four hundred microcoulomb. Clear? This sits on two plus three, that is five microfarad. So the common potential is four hundred over five, which gives eighty volts. Okay? Now the energy lost. In the loss formula, what goes in place of V one minus V two here? Unlike plates, so we replace V two by minus V two. V one minus the bracket minus V two becomes one hundred plus two hundred, which is three hundred volts. Right? Same capacitors, so one point two microfarad again. The loss is one half times one point two times ten to the power minus six, times three hundred squared, which gives zero point zero five four joules. Got it? Let us check with the energies. Before joining, the two energies, zero point zero one and zero point zero six joules, add up to zero point zero seven joules. After, one half times five microfarad times eighty squared gives zero point zero one six joules. The difference is zero point zero five four joules, a perfect match. Okay?

8:08Quick revision: energy, half as heat, sharing, plates apart

So, quick revision. First, U equals Q squared over two C, or one half C V squared, and of the battery's Q V, half becomes heat. Second, when capacitors are joined, charge is conserved but energy is lost; for unlike plates, use minus V two. And third, pulling plates apart, U rises if Q is fixed and falls if V is fixed. Now go and try the practice problems on this page. All the best!