C-1 Capacitors A capacitor is simply two pieces of metal near each other, separated by an insulator or air. A capacitor is used to store charge and energy. A parallel-plate capacitor consists of two parallel plates separated by a distance d, each plate with area A . If A is large and d is small, the plates are effectively infinite planes, and the E-field is uniform and entirely in-between the plates. L W +Q on top platehi V d d E area A = L ⋅W lo V –Q on top plate Charges are always on the inside surfaces, because (+) attracts (–) . The outside surfaces remain uncharged. "Charge Q on a capacitor" always means +Q on one plate, –Q on the other plate. Capacitors are charged by transferring (–) charge from one plate to the other. Taking (–) charge off a plate leaves behind an equal-sized (+) charge. The charges make an E-field, which means a voltage difference between the plates. The "voltage V on a capacitor" always means the voltage difference ∆V between the plates. Q| V | E d V V E Q ratio = constantV∆ = ⋅ = ⇒∝∝⇒ It is always true that ∆V ∝ E, since fiVEd∆=− ⋅∫rKK (double the E-field everywhere and ∆V doubles.) And it is always true the E ∝ Q, since ii2iiqˆEkr=∑Kr (double all the charges everywhere and E doubles). So the ratio Q/V is always a constant: if you double the charge Q, the V (=∆V) is guaranteed to double. Phys1120 Dubson 9/18/2009 ©University of Colorado at BoulderC-2 Definition: capacitance C of a capacitor: QCV= If we double the charge Q, the voltage V doubles, but the ratio Q/V remains constant. [Remember: Q means +Q and –Q , V means ∆V.] units [C] = coulomb / volt = farad (F) Big capacitance (1F) ⇒ can store a big Q with a small V Small capacitance (nF = 10–9 F) ⇒ small Q stored with a big V For a parallel-plate capacitor, with air or vacuum between the plates, the capacitance is oACdε= (air or vacuum separating plates) area Ad εo ("epsilon-naught") is the same constant that appeared in Gauss's Law. Proof: 00QdVV Ed dAσ∆= = ⋅= ⋅ = ⋅εε (We have used 0Eσ=ε for a capacitor.) Rearranging, we get oAQCVdε==. Done. Notice that the capacitance of a parallel-plate capacitor depends only on the size and shape of the two metal parts. This turns out to true of all capacitors. The capacitance of two pieces of metal depends solely on their geometry. Note that this formula means C increases as d decreases. Why? If Q is kept fixed, we have the same magnitude E-field (because same charge density σ = Q/A creates the E = σ/ε0 ). Smaller d and same-sized E means smaller voltage V = E d. Same Q and smaller V means bigger C = Q/V. smaller d E bigger C E Phys1120 Dubson 9/18/2009 ©University of Colorado at BoulderC-3 A farad is a huge capacitance. For example, suppose we make a parallel plate capacitor with area A = 1 m2 (big) and separation d = 1 mm = 0.001 m. The capacitance is only 129o3A(8.85 10 )(1)C 9 10 F 9nFd10ε−−−×== ×= (tiny!) Multi-farad capacitors in small packages are made by making d very small. d ≈ atomic dimensions ≤ nm (nanometer) is possible. Stored Energy in Capacitors It takes work to charge a capacitor, because it is difficult to transfer more electrons from the (+) plate to the (–) plate. The work required to transfer a charge q across a voltage difference "V" = ∆V is ∆PE = q ∆V. When we charge up a capacitor from qinitial = 0 to qfinal = Q, we transfer electrons one at a time. The first electron is easy to transfer since V = ∆V = 0 initially, but the later electrons take more and more work to transfer as Q (and ∆V) builds up. – e Total work to charge capacitor = electrostatic potential energy stored in capacitor = 1UQV2= (We use U for energy to avoid confusion with E for electric field.) (hard) q = – e (easy) Why the (1/2)? Why not ∆PE = Wext = Q V ? While transferring the total charge Q, the voltage difference increased from 0 to V. The average value was (1/2)V. We can show this more rigorously by doing an integral. When the voltage difference between the plates is V, the work required to transfer an extra bit of charge dq is dU = V dq = (q/C) dq . The total work (= total PE) to charge the capacitors is the sum (the integral) of the works done to transfer all the bits of charge: Q20qQUd U dqC2C== =∫∫Can rewrite U in various ways using C = Q / V, Q = C V, V = Q / C : Phys1120 Dubson 9/18/2009 ©University of Colorado at BoulderC-4 2211 1UQVCV22 2== =QC Where is this energy? The E-field contains the energy. It takes work to create an E-field. It turns out that the energy per volume (the energy density) of the E-field is given by 2oU1uEVol. 2ε== "Proof": ()N22200volumeA11 1UCV Ed E(A22d 2εε⎛⎞⎟⎜== = ⋅⎟⎜⎟⎜⎝⎠d) ⇒ 20U1uEVol. 2ε== (This is a proof for the special case of a parallel-plate capacitor only, but the result turns out to be true always.) The energy U = (1/2)QV of a charged capacitor is in the E-field between the plates. If we pull the plates apart, keeping the charge Q fixed, we increase the volume which contains E-field and the total energy increases. It was hard to pull the plates apart, because opposite charges attract. The work we did went into creating more E-field (same size E-field over larger volume). It turns out that the work done to assemble a collection of charges (Wext = ∆U = q ∆V) is equal to the energy in the E-field created: ()2102UEε= dV∫ (volume integral) Capacitors in parallel or in series. Symbol for capacitor : For capacitors in parallel, C|| = C1 + C2 E E = C||C1 C2 Phys1120 Dubson 9/18/2009 ©University of Colorado at BoulderC-5 A big C is equivalent to two smaller C side by side: "Proof" : For capacitors in series: series121C11CC=+ =C2 C1 Cseries Capacitors filled with dielectrics The capacitance C of a capacitor can be increased by placing an insulator ("dielectric") between plates. The dielectric is polarized by the charges on the plates. +Q polarized insulator –Q For fixed Q on the plates, the E-field between the plates is reduced when a dielectric is inserted because the polarization charge on the dielectric partially cancels the charge Q on the plates. smaller E ⇒ smaller V = |∆V| = E d , smaller V and same Q on plates ⇒ larger C = Q / V Let's call the original E-field E0 and the final, smaller, E-field E. The original E-field between the
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