lecture6 - 6. Lecture 6 6.1 Energy contained in a capacitor...

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Unformatted text preview: 6. Lecture 6 6.1 Energy contained in a capacitor We said that a capacitor stores energy. We now compute how much energy it actually contains. A way to do that is to consider how much work we need to separate the two plates that form the capacitor. Suppose the left plate is fixed and we move the right one by a distance ∆d (see fig.31). The electric field produced by the other plate is ￿ |E | = 2σ0 where σ = Q . The force is then ε A ￿ |F | = σ Q 2 ε0 (6.1) The work we need to do to separates the plates by a distance ∆d is σ ￿ W = ∆U = Ufinal − Uinitial = |F |∆d = Q∆ d 2 ε0 (6.2) From here we find that U= σ 1 d 1 Q2 Qd = Q2 = 2 ε0 2 Aε0 2C (6.3) where we remembered that the capacity is C= Aε0 d (6.4) Using that Q = C ∆V we find the equivalent expressions for the energy contained in the capacitor: 1 Q2 1 1 U= = C ( ∆ V ) 2 = Q∆ V (6.5) 2C 2 2 6.2 Dielectrics We mentioned before that between the plates in a capacitor we include an insulator. This can actually modify the properties of the capacitor and increase the capacity. The reason is that insulators are generically dielectric. This means that their electrical properties can be understood as if they are made out of tiny dipoles. The dipoles align themselves with the electric field decreasing its value in a phenomenon known as screening. If we consider a single charge inside a dielectric the screening phenomenon is illustrated in figure 32. Notice that outside the dielectric the field is the same from symmetry and Gauss law. Experimentally the result is that the electric field inside is – 40 – d ∆d V2 V1 Q E −Q Figure 31: The energy stored in a capacitor can be computed by doing a small displacement of the plates by ∆d and computing the work needed to do that. suppressed by a factor ε0 /ε where ￿ is known as the permitivity of the medium. The electric field inside the dielectric is then 1Q ￿ E= r ˆ 4πε r2 (6.6) outside is again 1Q r ˆ (6.7) 4πε0 r2 In the case of the capacitor it works the same, we just need to replace ε0 → ε. Therefore the capacity when there is a dielectric in between the plates is ￿ E= C= Aε d Typical values for ε/ε0 are ε Material ε0 Vacuum 1 (definition) Air 1.0006 Teflon 2.1 Polyethylene 2.25 Paper 3.5 Water 80 – 41 – (6.8) We see that air and vacuum are very close to each other which is why we ignore the difference. It also shows that if we want to store more energy in a capacitor, at the same voltage it is convenient to use a dielectric of high dielectric constant. + + − − + + + + + + − + − − + + + − − + − − − + + − − + + + − − − − − − − − + + + + − + + − + + − − − + + + − − + − + + + − − − + + − − + + − − + + − + − +q + + + + − − − − − − + − + + − − − + − + − + + + − − − + + + − − − − − − − + − + − + + + − + − + + − + + − Figure 32: A charge in a dielectric is partially screened by the dipoles aligning with the electric field. Due to temperature and other variables the alignments is partially random. As a result the field inside the dielectric is smaller that without the medium. Outside the medium, in this case, the electric field is the same with or without the dielectric. This is because the configuration is spherically symmetric and using Gauss’s law, the electric field is given by the total charge enclosed in a surface (the dielectric has no net charge). 6.3 RC circuits Now that we know what a capacitor is we can see how to charge it and what applications we can find for it. The simplest way to charge it is with a battery and through a resistor. One might think the resistor is not necessary but not matter what we do there is some resistance in the circuit so we need to include it. Also, sometimes we do not want to charge the capacitor as fast s we can and, as we are going to see, a resistor makes the charging process slower. When the contact is closed, current starts to circulate but it cannot go through the capacitor so charge starts to accumulate in it. Initially – 42 – the current is just I = ∆V but as the capacitor charges a potential difference appears R and the current is smaller. In the final stage, the capacitor is completely charged, the potential difference between its terminals is the same as the battery (but opposes it) and no more current circulates. An estimate of the time the capacitor takes to charge is obtained by dividing the total charge by the initial current: τ= Q C ∆V = = RC I ∆V /R (6.9) This is a characteristic time of the circuit. The larger the resistance and the capacity the larger the time it takes to charge. We can then discharge the capacitor through a smaller resistance for example and it will release its energy in a much shorter interval. An application is the flash in a photographic camera. The battery with relatively large internal resistance charges a capacitor and then it is suddenly discharged though the flash. Another typical application is in timing circuits. For example oscillators etc. A resistor of variable resistance is inserted and the time of charge and discharge is used to determine the period of the oscillations. It is not very precise so it can be used only when the exact timing is not crucial. Another important application is in computer memories. A charged capacitor for example can represent a 1 and a discharged one a 0. In that case we want the capacity small to be able to charge it and discharge it fast. However if we do not read the memory current will leak and the capacitor will be discharged. For that reason when the computer is turned off what was in the memory disappears. When it is working a special circuit continuously reads the memory and rewrites it before the capacitors have time to discharge (these memories are called DRAM and are presently the most common ones). We can analyze the circuit in slightly more detail by defining some points (a), (b), (c) as in figure 33 and considering the voltages at those points Va , Vb , Vc . Using Ohm’s law ∆V = IR for the resistor and Q = C ∆V for the capacitor we have Vb − Vc = ∆Vresistor = IR Q Va − Vb = ∆Vcapacitor = C Va − Vc = ∆Vbattery (6.10) (6.11) (6.12) If we add the first two we have Q = Va − Vc = ∆Vbattery C So the current circulating through the circuit is ￿ ￿ 1 Q I= ∆Vbattery − R C Vb − Vc + Va − Vb = IR + – 43 – (6.13) (6.14) Since ∆VBattery is fixed we see that initially, when the capacitor is not charged (Q = 0) 1 the current is I = R ∆VBattery . The current starts to charge the capacitor, Q increases and I decreases but keeps charging the capacitor until Q = C ∆Vbattery which gives I = 0. If we plot the voltage across the capacitor and the current we get a plot as in fig.34. C + + + + − − − − (a) I + (b) − (c) ∆V R Battery Figure 33: A capacitor is usually charged with a battery and through a resistor. The time it takes to charge depends on the capacity and the resistor. It is given by ∆t = RC and is a characteristic time of this type of circuit. – 44 – Figure 34: Time dependence of the voltage and current across the capacitor for the RC circuit in arbitrary units. It illustrates the general behavior of these quantities. – 45 – ...
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