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A Fully covered topic lecture for both computer Science Students and Electronic Students.
Typology: Cheat Sheet
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Figure 25-1 shows some of the many sizes and shapes of capacitors. Figure 25-2 shows the basic elements of any capacitor — two isolated conductors of any shape. No matter what their geometry, flat or not, we call these conductors plates. Figure 25-3a shows a less general but more conventional arrangement, called a parallel-plate capacitor, consisting of two parallel conducting plates of area A separated by a distance d. 2
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One way to charge a capacitor is to place it in an electric circuit with a battery. An electric circuit is a path through which charge can flow. A battery is a device that maintains a certain potential difference between its terminals (points at which charge can enter or leave the battery) by means of internal electrochemical reactions in which electric forces can move internal charge. In Fig. 25-4a, a battery B, a switch S, an uncharged capacitor C, and interconnecting wires form a circuit. The same circuit is shown in the schematic diagram of Fig. 25-4b, in which the symbols for a battery, a switch, and a capacitor represent those devices. The battery maintains potential difference V between its terminals. The terminal of higher potential is often called the positive terminal; the terminal of lower potential is often called the negative terminal. The circuit shown in Figs. 25-4a and b is said to be incomplete because switch S is open; that is, the switch does not electrically connect the wires attached to it. When the switch is closed, electrically connecting those wires, the circuit is complete and charge can then flow through the switch and the wires. As we know, the charge that can flow through a conductor, such as a wire, is that of electrons. 4
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When the circuit of Fig. 25-4 is completed, electrons are driven through the wires by an electric field that the battery sets up in the wires. The field drives electrons from capacitor plate h to the positive terminal of the battery; thus, plate h , losing electrons, becomes positively charged. The field drives just as many electrons from the negative terminal of the battery to capacitor plate l ; thus, plate l , gaining electrons, becomes negatively charged just as much as plate h , losing electrons, becomes positively charged. Initially, when the plates are uncharged, the potential difference between them is zero. As the plates become oppositely charged, that potential difference increases until it equals the potential difference V between the terminals of the battery. Then plate h and the positive terminal of the battery are at the same potential, and there is no longer an electric field in the wire between them. Similarly, plate l and the negative terminal reach the same potential, and there is then no electric field in the wire between them. Thus, with the field zero, there is no further drive of electrons. The capacitor is then said to be fully charged, with a potential difference V and charge q that are related by Eq. (1). Here, we assume that during the charging of a capacitor and afterward, charge cannot pass from one plate to the other across the gap separating them. Also, we assume that a capacitor can retain (or store) charge indefinitely, until it is put into a circuit where it can be discharged. 5
06/06/ Calculating the Potential Difference
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06/06/ Capacitance of a Parallel-Plate Capacitor
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06/06/ Capacitors in Parallel and in Series When there is a combination of capacitors in a circuit, we can replace that combination with an equivalent capacitor —that is, a single capacitor that has the same capacitance as the actual combination of capacitors. With such a replacement, we can simplify the circuit, affording easier solutions for unknown quantities of the circuit. Here we discuss two basic combinations of capacitors that allow such a replacement.
three capacitors are connected in parallel to battery B. This description has little to do with how the capacitor plates are drawn. Rather, “in parallel” means that the capacitors are directly wired together at one plate and directly wired together at the other plate, and that the same potential difference V is applied across the two groups of wired-together plates. Thus, each capacitor has the same potential difference V, which produces charge on the capacitor. (In Fig. 25-8a, the applied potential V is maintained by the battery.) In general, “When a potential difference V is applied across several capacitors connected in parallel, that potential difference V is applied across each capacitor. The total charge q stored on the capacitors is the sum of the charges stored on all the capacitors.” 10
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When we analyze a circuit of capacitors in parallel, we can simplify it with this mental replacement: Capacitors connected in parallel can be replaced with an equivalent capacitor that has the same total charge q and the same potential difference V as the actual capacitors. (You might remember this result with the word “par-V,” which means “capacitors in parallel have the same V.”) Figure 25-8b shows the equivalent capacitor (with equivalent capacitance C eq) that has replaced the three capacitors (with actual capacitances C 1 , C 2 , and C 3 ) of Fig. 25-8a. To derive an expression for C eq in Fig. 25-8b, we first use Eq. (1) to find the charge on each actual capacitor: q 1 =C 1 V, q 2 =C 2 V, and q 3 =C 3 V. The total charge on the parallel combination of Fig. 25-8a is then q=q 1 +q 2 + q 3 =(C 1 +C 2 +C 3 )V. The equivalent capacitance, with the same total charge q and applied potential difference V as the combination, is then Ceq ==C 1 +C 2 +C 3 A result that we can easily extend to any number n of capacitors, as Ceq= [n capacitors in parallel]. (10) Thus, to find the equivalent capacitance of a parallel combination, we simply add the individual capacitances. 11
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(^) We start with capacitor 3 and work upward to capacitor 1. When the battery is first connected to the series of capacitors, it produces charge -q on the bottom plate of capacitor 3. (^) That charge then repels negative charge from the top plate of capacitor 3 (leaving it with charge +q). The repelled negative charge moves to the bottom plate of capacitor 2 (giving it charge -q). (^) That charge on the bottom plate of capacitor 2 then repels negative charge from the top plate of capacitor 2 (leaving it with charge +q) to the bottom plate of capacitor 1 (giving it charge -q). (^) Finally, the charge on the bottom plate of capacitor 1 helps move negative charge from the top plate of capacitor 1 to the battery, leaving that top plate with charge +q. (^) Here are two important points about capacitors in series: (^) 1. When charge is shifted from one capacitor to another in a series of capacitors, it can move along only one route, such as from capacitor 3 to capacitor 2 in Fig. 25-9a. If there are additional routes, the capacitors are not in series. (^) 2. The battery directly produces charges on only the two plates to which it is connected (the bottom plate of capacitor 3 and the top plate of capacitor 1 in Fig. 25-9a). Charges that are produced on the other plates are due merely to the shifting of charge already there. (^) For example, in Fig. 25-9a, the part of the circuit enclosed by dashed lines is electrically isolated from the rest of the circuit. Thus, the net charge of that part cannot be changed by the battery— its charge can only be redistributed. 13
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(^) When we analyze a circuit of capacitors in series, we can simplify it with this mental replacement: (^) Capacitors that are connected in series can be replaced with an equivalent capacitor that has the same charge q and the same total potential difference V as the actual series capacitors. (^) (You might remember this with the word “seri-q” to mean “capacitors in series have the same q.”) Figure 25-9b shows the equivalent capacitor (with equivalent capacitance Ceq) that has replaced the three actual capacitors (with actual capacitances C 1 , C 2 , and C 3 ) of Fig. 25-9a. (^) To derive an expression for C eq in Fig. 25-9b, we first use Eq. (1) to find the potential difference of each actual capacitor: V 1 =V 2 = and V 3 = (^) The total potential difference V due to the battery is the sum of these three potential differences. Thus, V= V 1 +V 2 +V 3 =q(++) The equivalent capacitance is then = or =++ (^) We can easily extend this to any number n of capacitors as == [n capacitors in series]. (11) (^) Using Eq. (11) you can show that the equivalent capacitance of a series of capacitances is always less than the least capacitance in the series. 14
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(^) Solution: Any capacitors connected in series can be replaced with their equivalent capacitor, and any capacitors connected in parallel can be replaced with their equivalent capacitor. (^) Therefore, we should first check whether any of the capacitors in Fig. 25-10a are in parallel or series. Finding equivalent capacitance: Capacitors 1 and 3 are connected one after the other, but are they in series? No. (^) The potential V that is applied to the capacitors produces charge on the bottom plate of capacitor 3. That charge causes charge to shift from the top plate of capacitor 3. (^) However, note that the shifting charge can move to the bottom plates of both capacitor 1 and capacitor 2. Because there is more than one route for the shifting charge, capacitor 3 is not in series with capacitor 1 (or capacitor 2). (^) Are capacitor 1 and capacitor 2 in parallel? Yes. Their top plates are directly wired together and their bottom plates are directly wired together, and electric potential is applied between the top-plate pair and the bottom-plate pair. (^) Thus, capacitor 1 and capacitor 2 are in parallel, and Eq. (10) tells us that their equivalent capacitance C 12 is C 12 =C 1 +C 2 =12.0 μF+5.30 μF=17.3 μF. (^) In Fig. 25-10b, we have replaced capacitors 1 and 2 with their equivalent capacitor, called capacitor 12 (say “one 16
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(^) Is capacitor 12 in series with capacitor 3? Again applying the test for series capacitances, we see that the charge that shifts from the top plate of capacitor 3 must entirely go to the bottom plate of capacitor 12. (^) Thus, capacitor 12 and capacitor 3 are in series, and we can replace them with their equivalent C 123 (“one two three”), as shown in Fig. 25-10c. (^) From Eq. (11), we have =+ = +=0.280 F- (^) from which = (Answer) (^) (b) The potential difference applied to the input terminals in Fig. 25-10a is V=12.5 V. What is the charge on C 1? (^) Solution: We now need to work backwards from the equivalent capacitance to get the charge on a particular capacitor. We have two techniques for such “backwards work”: (^) (1) Seri-q: Series capacitors have the same charge as their equivalent capacitor. (2) Par-V: Parallel capacitors have the same potential difference as their equivalent capacitor. (^) Working backwards: To get the charge q 1 on capacitor 1, we work backwards to that capacitor, starting with the equivalent capacitor 123. Because the given potential difference V (=12.5 V) is applied across the actual combination of three capacitors in Fig. 25-10a, it is also applied across C 123 in Figs. 25-10d and e. 17
Work must be done by an external agent to charge a capacitor. Starting with an uncharged capacitor, for example, imagine that—using “magic tweezers”—you remove electrons from one plate and transfer them one at a time to the other plate. The electric field that builds up in the space between the plates has a direction that tends to oppose further transfer. Thus, as charge accumulates on the capacitor plates, you have to do increasingly larger amounts of work to transfer additional electrons. In practice, this work is done not by “magic tweezers” but by a battery, at the expense of its store of chemical energy. We visualize the work required to charge a capacitor as being stored in the form of electric potential energy U in the electric field between the plates. You can recover this energy, by discharging the capacitor in a circuit. Suppose that, at a given instant, a charge q’ has been transferred from one plate of a capacitor to the other. The potential difference V’ between the plates at that instant will be q’/C. If an extra increment of charge dq’ is then transferred, the increment of work required will be, from Eq. ∆V=V f -V i = , dW=V’dq’=dq’^19 06/06/
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The work required to bring the total capacitor charge up to a final value q is W=== This work is stored as potential energy U in the capacitor, so that U= (potential energy). (12) From Eq. (1), we can also write this as U= (potential energy). (13) Equations (12) and (13) hold no matter what the geometry of the capacitor is. In fact, this potential energy is stored in electric field between two plates. Therefore, we calculate potential energy in terms of electric field. As we know that C= and V=Ed. So by putting C and V in Eq. (13), we will obtain U=AE 2 d (14) “The potential energy of a charged capacitor may be viewed as being stored in the electric field between its plates”. In vacuum, the energy density u, or potential energy per unit volume, within an electric field of magnitude E is given by u=E 2 (15) Video link: https://youtu.be/XsLMwLcj3nY?t= 20
