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Chapter 26: Capacitance and Capacitors We can store electrical potential energy in device called a “capacitor”. Sometimes we think of capacitors as charge storage devices. Capacitors are in every electrical device you own or use. Capacitors range in size from µm to meters. The following example is a way to define capacitance (C) in terms of stored charge (q) and potential difference (V). Two metal plates, parallel to each other, each with area= A are separated by a distance=d. The top plate has +q on it, the bottom plate has –q. d The potential difference between the two plates is V. We define capacitance to be: C=q/V +q -q V Capacitance is a POSITIVE quantity Note: q is the charge on one plate (the sum of the charges on both plates is zero). In our example we defined capacitance using parallel conducting plates. We could have used concentric conducting spheres or cylinders. As we shall see, the value of capacitance only depends on the geometry of the plates (e.g. planar, spherical, cylindrical). By this we mean, if you know the size, shape and relative location of the plates you can calculate C without knowing q or V. The unit of capacitance is the farad. 1 farad = 1 F= 1 coulomb/volt Typical values (the size you might find in a TV, radio, CD player) range from 10-12 F to 10-6 F. Until recently (last 10 years or so) it was very rare to see a 1F capacitor. However, improvements in technology have made these capacitance values common place (and cheap!). This was largely driven by the computer industry. R. Kass P132 Sp04 1 Charging a Capacitor Suppose we take a battery and connect it to our parallel plate capacitor. - + battery +q -q V The battery will maintain a constant potential difference (V) between the two plates. For a brief time current will flow from the battery to the plates until the plates are fully charged. When the plates are fully charged there will be +q (=CV) on the upper plate and an equal amount of negative charge on the lower plate. Example: we use a 9V battery to charge a 1 µF capacitor: q=CV=(9 V)(10-6 F) = 9x10-6 C we would have +9x10-6 C on the top plate and - 9x10-6 C on the lower plate. As we will shortly be dealing with electrical circuits with batteries, capacitors and switches we can write a schematic of the electrical circuit using the following symbols: capacitor + - HRW Fig. 26-4 switch open position battery R. Kass P132 Sp04 2 Calculating Capacitance Let’s calculate the capacitance of our parallel plate capacitor. area of plate=A d We want a formula that only depends on geometrical factors, e.g. area of plates (A), separation of plates (d). E +q -q In ch. 24, section 8 the electric field in between two conducting parallel plates was discussed. Away from the edges of the plates the E-field is constant and perpendicular to the plate as shown in the figure. Gauss’ Law gives: ∫ E ⋅ dA = q enc /ε0 ∫ E ⋅ dA = EA = q enc q= charge on one plate /ε0 ⇒ E = conductor E=0 ds q s=d s=0 E ε0A To calculate the capacitance we need to know the potential difference between the plates: d d ∫ d ∫ ∫ Since capacitance is a positive quantity we don’t care about the sign of V. V = − E ⋅ ds = − Eds cos(180) =E ds =Ed 0 0 0 Using the value of the E-field we obtained from Gauss’ Law we find: V = Ed = qd ε0A ε0 = 8.85x10-12 F/m Finally, using the definition of capacitance, C=q/V, we find: C= ε A q q = ⇒C = 0 qd V d ε0A Note: our formula for C only contains geometric factors, area and distance. To make a 1F capacitor out of square parallel plates in air separated by 1mm the plates have to be ~10 km per side !! R. Kass P132 Sp04 3 Calculating Capacitance continued Let’s do one more example, a spherical capacitor made of two concentric spheres. -q First, we calculate the electric field in the region between the spheres. The inner sphere has +q and the outer sphere has –q distributed on it. We get a familiar result using Gauss’ Law: ∫ E ⋅ dA = q enc ∫ 2 / ε 0 ⇒ E dA = E (4πr ) = q ε0 ⇒E= q a ∫ a ∫ ∫ Vb − V a = − E ⋅ ds = − E cos(180)dr = Edr = b b b a q 4πε 0 dr ∫r 2 =− b q 4πε 0 r a b +q r 4πε 0 r 2 Next, we need the potential difference (we’ve done this before too..): a ra E q 1 1 =− [ − ] 4πε 0 a b rb Since capacitance is a positive quantity, we don’t care about the sign of V. So, we can take its absolute value and call it V (=|Vb-Va|). q 1 1 q b−a V =| Vb − V a |= [ − ]= 4πε 0 a b 4πε 0 ab The capacitance is just C=q/V: C= q = V q b−a 4πε 0 ab q ⇒C= 4πε 0 ab b−a Again, the capacitance only depends on geometric factors, the inner and outer radii this time. R. Kass P132 Sp04 4 Combining Capacitors in Parallel Capacitors are very common circuit elements. Often a circuit will contain more than one capacitor. As we will see, two (or more) capacitors can be can be connected in a fashion that either increases or decreases the total capacitance. First, consider the case where the capacitors are combined in PARALLEL. Here we take a battery and connect the positive battery terminal to one end of each capacitor and the negative terminal to the other end of the capacitor. We want to find the equivalent capacitance of the 3 parallel capacitors. Since the capacitors are connected in parallel they all have the same potential difference. V=q1/C1=q2/C2=q3/C3 We can rearrange the above equation in terms of the total charge qtotal=q1+q2+q3 qtotal=VC1+VC2+VC3=V(C1+C2+C3) We can rewrite the above in terms of an equivalent capacitance Ceq: Capacitors in parallel have an equivalent qtotal=VCeq capacitance that is larger than the value of each capacitor! Ceq=C1+C2+C3 HRW Fig. 26-7 Thus n capacitors in parallel have equivalent capacitance: n Ceq = ∑C i i =1 R. Kass P132 Sp04 5 Combining Capacitors in Series The other way to combine capacitors is in SERIES. Here each capacitor is wired serially and therefore the potential difference across each capacitor can be different. In this configuration each capacitor has the same electric charge on its plates. q=C1V1=C2V2=C3V3 The sum of the voltage differences has to add up to the battery voltage (why?) V=V1+V2+V3 V=q/C1+q/C2+q/C3 We can rewrite this as: 1 q 1 1 ⇒ V = q + + Ceq C1 C 2 C3 Here we identify Ceq as: 1 1 1 1 = + + Ceq C1 C 2 C3 For n capacitors in series we have: 1 = Ceq R. Kass n 1 ∑C i =1 i P132 Sp04 Capacitors in series have an equivalent capacitance that is smaller than the value of each capacitor! 6 Combining Capacitors In an electrical circuit capacitors are not always strictly in parallel or series. In these 4 circuits C1 and C2 are always in parallel. However, C3 is not in parallel or in series with either of these capacitors. C3 is in series with the parallel equivalent of C1 and C2. For example, assume C1=C2=2µF and C3=1µF. The parallel equivalent of C1 and C2 is: Ceq=C1+C2=(2µF)+(2µF)=4µF The circuit behaves as if a 4µF capacitor was in series with C3. The equivalent capacitance (Ceq total) of the combination of the 3 capacitors is: 1 1 1 1 1 5 = + = + = Ceq total Ceq1, 2 C3 4 1 4 The equivalent capacitance (Ceq total) of the combination of the 3 capacitors is: Ceq total=4/5 µF Thus we could have replaced the 3 capacitors with one 4/5 µF capacitor and the circuit would be behave identically. R. Kass P132 Sp04 7 Energy Stored in a Capacitor A capacitor is an energy storage device. We can calculate the energy stored in a capacitor by calculating the work done to put electric charges on the plates of the capacitor. The amount of work (dW) that it takes to put some charge (dq) on the plates of a capacitor with potential difference V between its two plates is: dW=Vdq (for this derivation we don’t care about the sign of W) We can rewrite this equation in terms of dq eliminating V using: q C is a constant dq C The total work to put all of the charge on the capacitor is just the integral of dW: dW = Vdq = q2 1 W= qdq = C 2C ∫ This energy is stored as potential energy in the electric field of the capacitor. Therefore the energy in a capacitor is: q2 E= 2C 2 CV We can rewrite the above equation in terms of V instead of charge q: E = 2 Example: a 200µF capacitor held at 5000V stores 2500 Joules of energy! R. Kass P132 Sp04 8