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Capacitors We will use a parallel plate capacitor for our discussions here. We already know that the electric field between the plates is a uniform field with equipotentials parallel to the plates and equally spaced. The lines of force are perpendicular to these, running straight from the positive plate to the negative one. When there is a potential difference of ΔV between the plates, the potential and electric field can be mathematically represented as V = ΔV x/d where x is the distance from the negative plate. E = ΔV/d anywhere between the plates and directed from the positive to the negative plate. We need to relate these to the charge on the plates. The charge density on the plates, σ = q/A, increases with increasing potential, ΔV, and with decreasing plate separation, d. Both these factors strengthen the field and provide the force necessary to pack the charges together closer to increase the charge density. So, we have σ = q/A ΔV/d Rearranging gives q = constant(AΔV/d) It turns out that the amount of charge on the plate is also affected by the material between the plates. Materials respond to electric fields in different ways. A conductor placed in the field becomes a dipole with charges accumulating at each end. Insulating materials respond differently. To the extent the material allows charges to move, they respond similarly to conductors, but to a lesser degree. Some of these materials, also called dielectrics, have molecules with positive and negative ends, called dipoles. Water is a good example of this. These molecules can align with the field to modify the effective field strength in the material. When the dielectric material has electric dipoles which can align, they strengthen the field between the plates, increasing the charge stored on the plates. The parameter used to model this effect is the dielectric constant, . This factor multiplies the permittivity of free space, o, which is 4 divided by the constant in the Coulomb force law: o = 4/K = 4/9E9 Putting this factor in as the constant in the capacitor equation, q = o (AΔV/d) Rearranging, q = (o A/d) ΔV = C ΔV The collection of variables called C above all involve characteristics of the parallel plate capacitor. (o A/d) = C o A d : property of the material between the plates. : Area of a plate of the capacitor. : the distance between the plates. A physical capacitor, once constructed, will have a capacitance determined by the geometry and choice of dielectric used to make the capacitor. Commercial capacitors are labeled with the capacitance value. The value of the dielectric constant, , of a material is often determined by placing a measured amount of the material between the plates of a capacitor and measuring its capacitance. Energy: Storing charge in a capacitor also stores energy. The charge on the plates has a potential energy due to the repulsion between charges. This potential energy is stored in the electric field. The amount of potential energy can be determined by considering charge to have been moved from one plate to the other through the potential difference ΔV. As charge is separated, the potential difference grows from 0 to ΔV. To figure the energy, use the average potential difference, ΔV/2, and the total charge, Q, to give the total potential energy as PE = Q ΔV/2 = ½ Q2/C = ½ C (ΔV)2 or ½ CV2 The second and third forms are obtained using the capacitor equation Q = C ΔV to substitute for either Q or ΔV. The potential difference, ΔV, is often written simply as V. Adding capacitors Two capacitors can be put in a circuit in two different ways; side-by-side (parallel) and in line (series). Capacitors wired in parallel effectively increase the area of the plates, adding the two areas. This gives a total capacitance equal to the sum of the individual capacitances. Ctot = C1 + C2 Wired in series, the simplest approach is to notice that the connection between the capacitors, along with the plates connected is completely insulated by the dielectrics, so the total charge must remain 0. This means that –Q on the plate of one capacitor is the same amount of charge as +Q on the plate of the other capacitor. This means that the amount of charge on each capacitor is the same. We also know that the total potential difference across both capacitors is equal to the sum of the potential differences across each capacitor. Using the capacitor equation and putting it all together: ΔV1 = Q/C1 ΔV2 = Q/C2 ΔV = Q/Ctot (same Q on outer plates), Q/C1 + Q/C2 = 1/C1 + 1/C2 = Q/Ctot 1/Ctot