so E hasto be Q/(2A epsilon naught) However, it seems rather inconsistentin the case of the parallel plate capacitor. Sign in with Google Sign in with Microsoft. I know the formula says E= charge density/epsilon naught. One plate gets +Q/2 and the other gets -Q/2 Is it because we are looking at the mid section of the conductor assome kind of Faraday cage? (since the plate is essentially infinityand blocks the electric field from the other side?)Īlso, lets look at two parallel plate capacitor. Then outside the space between the two plates, E = 2*(Q/4A*epsilonnaught)īut why cant we use superposition principle in the case of oneconducting plate with Q and say you get E=Q/(A epsilon naught)i.e.Q/(2A epsilon naught) from each surface. Physics 211 Exams 1) 2) 3) Practice Final Exam (A) The next three questions.
Please sign up for a Final Exam slot at CBTF if you have not already done so. Course Information: Credit is not given for both PHYS 211 and PHYS 101. Then, if you have two nonconducting plates separated by distance Dand each plate gets Q/2 which is uniformly distributed. PHYS 101 :: Physics Illinois :: University of Illinois at Urbana-Champaign. The charge will be equally split into two and concentrated only onthe two surfaces.Īnd the magnitude of electric field outside the plate isE=Q/(2A*epsilon naught) (E for a conducting plate is surface chargedensity/epsilon naught as opposed to surface chargedensity/(2epsilon naught) Lets say you have an infinite conducting plate with finitethickness D and it is given charge Q You have ninety (90) minutes to complete it: use a 2 pencil do not use a mechanical pencil or a pen.
Document Summary Turn off your cell phone and put it away. This has been bothering me for a long time and I cant find anadequate explanation. University of Illinois Department Physics Course PHYS 101 Professor For unlimited access to Study Guides, a Grade+ subscription is required.
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