ps04sol

# ps04sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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PS04 Solution - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 4 Solutions Problem 1: Parallel Plate Capacitor A parallel-plate capacitor is charged to a potential V 0 and charge Q 0 . It is then disconnected from the battery. The separation of the plates is then halved. What happens to: (a) the charge on the plates? No Change. We aren’t attached to a battery, so the charge is fixed. (b) the electric field? No Change. The charge is constant so, in the planar geometry, so is the field. (c) the energy stored in the electric field? Halves. The volume in which the field exists, halves, so the energy does too. (d) the potential? Halves. V = E d , so if d halves, so does V (e) How much work did you do in halving the distance between the plates? The work done is the change in energy. Energy, given the charge and potential, is: 2 11 22 UC V Q V == The energy halves, so the change is half the initial energy: 1 00 4 WU Q V =∆ =− Notice the sign – you did negative work bringing the plates together because that is the way they naturally want to move; the field did positive work. Problem 2: Capacitor A parallel plate capacitor has capacitance C . It is connected to a battery of voltage V until fully charged, and then disconnected. The plates are then pulled apart an extra distance d , during which the measured potential difference between them changed by a factor of 2. Below are a series of questions about how other quantities changed. Although they are related you do not need to rely on the answers to early questions in order to correctly answer the later ones. a) Did the potential difference increase or decrease by a factor of 2? DECREASE

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PS04 Solution - 2 Problem 2: Capacitor continued b) By what factor did the electric field change due to this increase in distance? Make sure that you indicate whether the field increased or decreased. Since the charge cannot change (the battery is disconnected) the electric field cannot change either. c) By what factor did the energy stored in the electric field change? Make sure that you indicate whether the energy increased or decreased. The electric field is constant but the volume in which the field exists increased, so the energy must have increased. But by how much? The energy 1 2 UQ V = . The charge doesn’t change, the potential increased by a factor of 2, so the energy: Increased by a factor of 2 d) A dielectric of dielectric constant κ is now inserted to completely fill the volume between the plates. Now by what factor does the energy stored in the electric field change? Does it increase or decrease? Inserting a dielectric decreases the electric field by a factor of κ so it decreases the potential by a factor of κ as well. So now, by using the same energy formula 1 2 V = , Energy decreases by a factor of κ e) What is the volume of the dielectric necessary to fill the region between the plates?
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## This note was uploaded on 02/22/2009 for the course 8 02 taught by Professor Sciolla during the Fall '08 term at MIT.

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ps04sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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