Download Alternating Current - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4 (30 pts) Two large, parallel conducting plates of area 40 cm2 and positioned 2 cm apart are connected to a source of alternating current. At one instant in time, the left plate carries a charge of 15 µC while the right plate has a charge of -15 µC. a) (15 pts) Compute the electric potential difference between the two plates. There are at least two ways to approach this problem. I. Since we know that the electric field between the parallel plates of a capacitor is σ/ǫ0 (or you can work it out with Gauss’s Law)∗, we can find the electric field E = 15 × 10−6 8.854× 10−12 = 4.24 × 108 V/m Then, compute the electric potential along a seperation of 2 cm: |∆V | = ∫ ~E · d~ℓ = ( 4.24 × 108 ) (0.02) = 8.5 × 106 V II. Recalling that the capacitance of a parallel plate capacitor is∗∗ C = ǫ0A d = ( 8.854× 10−12 ) (0.004) 0.02 = 1.77 × 10−12 F The definition of capacitance gives C = Q V which can be solved for V giving V = Q C = 15 × 10−6 1.77 × 10−12 = 8.5 × 106 V ∗ This has appeared in homework and on exams this semester as well as in the examples of past exams. For example, see Problem 1(b) on Exam 1. Homework problem 19.41 also deals with the electric field near a large flat sheet, as does Example 19.12 in the text. If you had forgotten this result, it is easy to reproduce it. Start with a single infinite flat sheet of charge with charge density σ and apply Gauss’s Law. A σ EE The sides of the “cookie-cutter” shaped Gaussian surface do not contribute to the flux integral since the electric field is always pointed directly away from the sheet (true on both sides). Only the two end faces, shown as area A in the figure above, contribute to the flux. This gives 2AE = σA ǫ0 | ~E| = σ 2ǫ0 Of course, this is the field around only a single infinite sheet. We need the field between two infinite sheets of opposite charge, as shown in the figure below. Between the sheets, the contribution from the fields from both sheets are in the same direction (away from the positive sheet and towards the negative sheet). Thus between the sheets the total field is | ~E| = 2 ( σ 2ǫ0 ) = σ ǫ0 The field outside of the plates is zero since the contributions to the total field from each of the sheets are in opposite directions but are of equal magnitude. E1 E2 E1 E2 E2E2 E2 E2 E1 E1E1 E1 E = 0 E = 2E E = 0 −σ+σ sheet 1 sheet 2 ∗∗ If you had forgotten how to compute the capacitance of a parallel plate capacitor from geometrical considerations, it can be quickly derived. If we assume the capacitor has a charge Q, the surface charge density on the plates are σ = ±Q/A where A is the area of each plate rather than of the end faces of the Gaussian surface. The derivation above gives the electric field inside the capacitor as | ~E| = σ/ǫ0. Then we can compute the potential difference between the two plates with this electric field, |∆V | = ∫ ~E · d~ℓ = E d = σd ǫ0 Plugging the potential between the plates into the definition of capacitance gives C = Q V = ǫ0σA σd = ǫ0A d This is a prime example where a thorough knowledge and understanding of the basic principles (here Gauss’s Law and the definition of capacitance) can quickly lead to whatever special case equations are necessary for a particular problem without the need to try to memorize hundreds of special equations in addition to the specifics of the problems to which they may be applied.
