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lecture6 - 6 Lecture 6 6.1 Energy contained in a capacitor...

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6. Lecture 6 6.1 Energy contained in a capacitor We said that a capacitor stores energy. We now compute how much energy it actually contains. A way to do that is to consider how much work we need to separate the two plates that form the capacitor. Suppose the left plate is fixed and we move the right one by a distance d (see fig.31). The electric field produced by the other plate is | E | = σ 2 ε 0 where σ = Q A . The force is then | F | = σ 2 ε 0 Q (6.1) The work we need to do to separates the plates by a distance d is W = U = U final U initial = | F | d = σ 2 ε 0 Q d (6.2) From here we find that U = σ 2 ε 0 Qd = 1 2 Q 2 d A ε 0 = 1 2 Q 2 C (6.3) where we remembered that the capacity is C = A ε 0 d (6.4) Using that Q = C V we find the equivalent expressions for the energy contained in the capacitor: U = 1 2 Q 2 C = 1 2 C ( V ) 2 = 1 2 Q V (6.5) 6.2 Dielectrics We mentioned before that between the plates in a capacitor we include an insulator. This can actually modify the properties of the capacitor and increase the capacity. The reason is that insulators are generically dielectric. This means that their electrical properties can be understood as if they are made out of tiny dipoles. The dipoles align themselves with the electric field decreasing its value in a phenomenon known as screening. If we consider a single charge inside a dielectric the screening phenomenon is illustrated in figure 32. Notice that outside the dielectric the field is the same from symmetry and Gauss law. Experimentally the result is that the electric field inside is – 40 –

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V 1 V 2 d d Δ Q - Q E Figure 31: The energy stored in a capacitor can be computed by doing a small displacement of the plates by d and computing the work needed to do that.
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