PHY2049ch25C%282-10-10%29

PHY2049ch25C%282-10-10%29 - HRW uses Gausss law & +...

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To find for the cylindrical capacitor of inner conductor radius a , outer conductor radius b and length L the capacitance, a b Cylinder o L C2 ln(b/a) ε While for a spherical capacitor of inner conductor radius a , outer conductor radius b , a capacitance, Sphere o o ab a C4 4 a ba 1 b ε ε HRW uses Gauss’s law & surf surf VV V E d s + +− =−= G G
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isolated o o b Sphere R Cl i m 4 4 R R 1 b →∞ ε = π ε The form for the spherical capacitor is useful because, if we call the radius of the inner sphere R , and take the limit as the outer conductor radius, b, goes to infinity we get the capacitance of an isolated sphere , With we can revisit the problem of electrically connecting charged metal spheres and how the charge redistributes between them. o C4 R ε Initially we required that the spheres have the same radius in which case the charge on each sphere becomes the numerical average of the two initial charges. Q 1 Q 2 12 QQ 2 + ′′ == before after far apart 1 Q 2 Q
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Q 1 Q 2 before after Upon learning that metals are equipotential surfaces and using the expression for the potential of a spherical charge distribution, far apart 1 Q 2 Q We now treat this case by the capacitance of isolated (since they are far apart) spheres. Something we demand of any theory is internal consistency, so these kinds of checks are useful. you learned how to handle charge redistribution between metal spheres of different radii. o 1Q V 4R = πε 1 V 2 V 1 V 2 V
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When the metals are electrically connected they electrically equilibrate (by exchanging charge) to become an equipotential. So after the wire connects them we must have that, 12 QQ CC = VV = Since we can write, so the equality of the two potentials implies that, qC V = Q 1 Q 2 before after far apart 1 Q 2 Q 1 V 2 V 1 V 2 V q V C =
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And since for an isolated sphere we have, 12 o1 o2 QQ 4R = πε πε o C4 R = πε RR = There is one more requirement, which is that the total charge does not change, i.e., Q + = Solving for : and using this in the * expression, Q = (where ) Q = + 1 Q 22 Q ′′ = 2 2 1 QR = * 1 2 2 QR QR = + 1 2 QR Q (R R ) = +
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and similarly, 1 1 12 R QQ RR = + Note that If R 1 = R 2 = R , where Q =+ R 1 R R 2 = == ++ + Our previous result for identical spheres . In which case 1Q Q Q 22 + ′′ = Which gives the new charge on each sphere for the general case (where the radii can differ). 2 2 R = + We find,
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We can use the general result to quantify how effective connection to earth ground is at discharging a charged object Suppose we have a metal sphere with a radius of 10 cm (0.1 m) and we charge it up to 10,000 V.
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PHY2049ch25C%282-10-10%29 - HRW uses Gausss law & +...

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