Chapter 25 Capacitance-II In the last lecture: we calculated the capacitance C of a system of two isolated conductors. We also calculated the capacitance for some simple geometries. In this chapter we will cover the following topics: -Methods of connecting capacitors (in series, in parallel). -Equivalent capacitance. -Energy stored in a capacitor. -Behavior of an insulator (a.k.a. dielectric) when placed in the electric field created in the space between the plates of a capacitor. -Gauss’ law in the presence of dielectrics. (25 - 1)
HITT A. the work done by the field is positive and the potential energy of the electron-field system increases B. the work done by the field is negative and the potential energy of the electron-field system increases C. the work done by the field is positive and the potential energy of the electron-field system decreases D. the work done by the field is negative and the potential energy of the electron-field system decreases E. the work done by the field is positive and the potential energy of the electron-field system does not change An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement: i j
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A question What is the equivalent capacitance between the points A and B? A.1 μF B. 2 μF C.4 μF D.10μF E.None of these AB What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC
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HITT A parallel-plate capacitor has a plate area of 0.2m 2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10 6 V/m between the plates, the magnitude of the charge on each plate should be: A. 8.9 x CB. 1.8 x C C. 3.5 x CD. 7.1 x C E. 1.4 x C
A question Each of the four capacitors shown is 500 μ F. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is: A. 0.2B. 0.5C. 20D. 50E. none of these
Question A parallel-plate capacitor has a plate area of 0.3m 2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10 -6 C then the force exerted by one plate on the other has a magnitude of about: A. 0B. 5NC. 9ND. 1 x10 4 N E. 9 x 10 5 N
Question A parallel-plate capacitor has a plate area of 0.3m 2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10 -6 C then the force exerted by one plate on the other has a magnitude of about: A.0 B.5N C.0. 9N D.1 x10 4 N E.9 x 10 5 N The electric field = σ/2ε o why?