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How Does Gauss' Law Apply to a Spherical Conducting Shell with a Central Charge?
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[QUOTE="Ignitia, post: 5937304, member: 640012"] [h2]Homework Statement [/h2] A point charge q=−5.0×10−12 C is placed at the center of a spherical conducting shell of inner radius 3.5cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b)What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?[h2]Homework Equations[/h2] E=K (Q/r[SUP]2[/SUP]) K: Coloumb's Constant, so (E*r[SUP]2[/SUP])/K = Q with Q being the derivative of the enclosed q=−5.0×10−12 C and σ= Q/S With S = ∫ 4πr[SUP]2[/SUP]dr [B] 3. The Attempt at a Solution [/B] So if I plug in the values accordingly, I should get: a) r = 3.5 σ= Q/S σ= ((E*r[SUP]2[/SUP])/K)/∫ 4πr[SUP]2[/SUP]dr (integrate from 0 to 3.5) [LEFT][SIZE=4][COLOR=rgb(5, 5, 5)] b) r = 4.5 σ= Q/S σ= ((E*r[SUP]2[/SUP])/K)/∫ 4πr[SUP]2[/SUP]dr (integrate from 3.5 to 4.5) c) Should just be the same as q=−5.0×10−12 C Before I plug this into a calculator and get the value, is this correct? Edit: The integrate sign isn't showing up, so the little circle is actually that.[/COLOR][/SIZE][/LEFT] [/QUOTE]
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How Does Gauss' Law Apply to a Spherical Conducting Shell with a Central Charge?
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