Lec7 - Physics 8B Professor Catherine Bordel 09/15/10...

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Physics 8B Professor Catherine Bordel 09/15/10 Lecture 8 ASUC Lecture Notes Online is the only authorized note-taking service at UC Berkeley. Do not share, copy, or illegally distribute (electronically or otherwise) these notes. Our student-run program depends on your individual subscription for its continued existence. These notes are copyrighted by the University of California and are for your personal use only. D O N O T C O P Y Sharing or copying these notes is illegal and could end note taking for this course. LECTURE We’re going to start with the problem of the spherical capacitor. You remember we have a perfect sphere with radius r 1 , surrounded by a second conductor of inner radius r 2 . The inner sphere has charge +q while the inner surface of the outside sphere has charge –q, and you were supposed to find the capacitance. We had determined that the electric field is radial and points outwards because you have a positive charge +q on the inner sphere. Make sure you know how to make the drawing at any possible point. I remind you that the definition of capacitance is C = Q/U. U is the difference in electric potential between the two electrodes. Each of the surfaces is equipotential. We need to calculate that potential difference V1 – V2. 1 2 2 1 0 2 1 0 1 2 1 2 1 2 0 2 1 0 2 1 0 2 1 2 0 2 1 4 4 ) ( 4 1 1 4 1 4 4 2 1 2 1 2 1 2 1 R R R R C R R R R Q Q C R R R R Q U R R Q U V V r Q V V dr r Q dV l d E V V Q U Q C R R R R R R V V   Student : Where did the negative sign go? Initially it was V 2 -V 1 , but here we have V 1 –V 2 which takes the negative sign into account. Student : Can you explain why r 2 became r?
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This note was uploaded on 10/02/2010 for the course PHYSICS 8B taught by Professor Shapiro during the Spring '07 term at University of California, Berkeley.

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Lec7 - Physics 8B Professor Catherine Bordel 09/15/10...

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