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Capacitance Chapter 26 (Continued) Capacitors Q + -Q V - If the plates are held at an electric potential difference V then get charges Q and -Q on the plates with Q=CV V should be really be written ∆V, but we often don’t bother. The battery’s ability to push charge is called its “electromotive force” or emf. A 6V battery has an emf of 6V. We often refer to electric potential, potential difference, and emf simply and sloppily as “voltage,” because all have units of volts. Dielectrics in Capacitors • Suppose we fill the space between the plates of a capacitor with an insulating material (a “dielectric”): +Q -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -Q • The material will be “polarized” - charge -e electrons are pulled to the left, stretched away from the positively charged atomic cores. • Consequently the E field within the capacitor will be reduced. Dielectrics in Capacitors +Q -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -Q • At a point in the interior some electrons shift left. But other electrons shift over to replace them (on average). So there is no net charge density in the bulk. • However, some electrons are pushed towards the left plate, and there is a deficit of electrons at the right plate. • Hence there is some induced charge -Q’ at the very left side of the insulator, and an induced charge +Q’ at the very right side. Dielectrics in Capacitors +Q + + + + + + + + + - -Q’ + + +Q’ -+ + - -Q It turns out that Q’ is proportional to Q. It is convenient to write the constant of proportionality this way: 1 Q' Q is called the dielectric constant. Its value depends on the dielectric (whether it is plastic, glass, etc.) For a vacuum =1; for air is close to 1. Except for a vacuum >1. Dielectrics in Capacitors +Q Gaussian Surface (a 3D box) + + + + + + + + + - -Q’ + + +Q’ -+ + - -Q Suppose the field inside is E0 without the dielectric. Let us calculate the new field E when the dielectric is present. Use Gauss’s Law in both cases: E0 = Q / (0 A) and E = (Q – Q’) / (0 A) E0 / E = Q / (Q – Q’) E0 / E = or E = E0/ Thus the field is reduced by factor Dielectrics in Capacitors • We see that a dielectric reduces the electric field by a factor (E=E0/) • Hence for a given charge Q on the metal plate the potential difference V = Ed is also reduced by (V=V0/). • Thus C= Q/V is increased by [C=C0 where C0=Q/V0 is the capacitance without the dielectric]. C oA d parallel plate capacitor with dielectric. • Adding a dielectric increases the capacitance. Dielectrics & Gauss’s Law With a dielectric present, Gauss’s Law can be rewritten from +q to 0 E dA q q' -q’ 0 E dA q Instead of having to think about the confusing induced charge q’, we can simply use the free charge q. But E is replaced by E. Parallel and Series Parallel Series Capacitors in Circuits +Q -Q (Symbol for a capacitor) C V A piece of metal in equilibrium has a constant value of potential. Thus, the potential of a plate and attached wire is the same. The potential difference between the ends of the wires is V, the same as the potential difference between the plates. Capacitors in Parallel • Suppose there is a potential difference V between a and b. • Then q1=C1V & q2=C2V • We want to replace C1 and C2 with an equivalent capacitance for which q=CV, where the charge on C is q = q1 + q2 a q1 q2 C1 b C2 V a C-q • From the above, q=q1 + q2=(C1 + C2)V. That is, q=CV with C = C1 + C2 • This is the equation for capacitors in parallel. • Increasing the number of capacitors increases the capacitance. b Capacitors in Series a C1 C2 -q +q -q +q V1 C b a -q V2 +q b V • The capacitors have the same q but different potential differences. Here the total potential difference between a and b is V = V1 + V2 • Again we want to find q=CV with C an equivalent capacitance. • Use V1 = q/C1 and V2 = q/C2 • Then V = V1 + V2 = q /C1 +q /C2 = [(1/C1) + (1/C2)]q • or, V = q/C with 1 / C = 1 / C1 + 1 / C2 • This is the equation for capacitors in series. • Increasing the number of capacitors decreases the capacitance. Equivalent Capacitors Parallel C1 C2 C = C1 + C2 Series For each there is an equivalent single capacitor C for which q=CV. Here q is the total charge to flow down the main wire, and V the potential difference across the unit. C1 C2 1 / C = 1 / C1 + 1 / C2