734_Physics ProblemsTechnical Physics

734_Physics ProblemsTechnical Physics - 74 Electric...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
74 Electric Potential P25.70 Inside the sphere, EEE xyz === 0 . Outside, E V xx VE zE a z x yz x =− −+ + + F H I K 00 0 32 22 32 ej So E E a z x E a x z x =− + + F H G I K J ++ L N M O Q P =+ + −− 3 2 23 0 2 2 52 0 2 2 a f E V yy a z x y z EE a z x y z yE a y z x y z E V z a z axyz E a z x y x y y y z z + + F H I K F H G I K J = F H G I K J + 0 0 2 2 0 2 2 2 2 0 32 2 2 32222 3 2 3 2 2 2 a f 22 + z P25.71 For an element of area which is a ring of radius r and width dr , dV kdq rx e = + . dq dA Cr rdr == σπ 2 bg and VC k rd r CkRR x x x RRx e R e = + + F H G I K J L N M M O Q P P z 2 2 0 222 ππ b g ln . P25.72 dU Vdq = where the potential V kq r e = . The element of charge in a shell is dq = ρ (volume element) or dq r dr = ρπ 4 2 and the charge q in a sphere of radius r is qr d r r r F H G I K J z 4 4 3 2 0 3 πρ π . Substituting this into the expression for dU , we have dU r dq k r r r k r Ud U k r d r k R e ee e R e = F H G I K J = F
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online