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Electric & Magnetic Fields

by: Jalon Willms

Electric & Magnetic Fields PH 316

Jalon Willms
GPA 3.69


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This 4 page Class Notes was uploaded by Jalon Willms on Monday October 19, 2015. The Class Notes belongs to PH 316 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/225116/ph-316-rose-hulman-institute-of-technology in Physics 2 at Rose-Hulman Institute of Technology.


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Date Created: 10/19/15
PH 316 Images with the law of cosines MJM September 17 2006 First we arrange two charges q and P q 0cq 1 1 1 2 R so that the pair produce a spherical region of 9 V 0 equivalent to a grounded conducting q 0cq sphere of radius R l4 s gt lt s gt The sketch is badly drawn since R will turn out to be between s and s From the law of cosines r12R2s22Rs cos 9 and 1 r22 R2 s 2 2Rs cos 9 2 Now we let ss R2 so that s Rzs Then we substitute in 2 for s r22 R2 Rzs2 2RR2s cos a R2 R4s2 2 R3s cos a 3 Next we sh out Rzs2 from Eq 3 r22 ms2 R2 s2 2Rs cos 9 4 But the expression in curly brackets is just r12 so that we wind up with r IVs r1 5 The potential at point p is Vp kqr1 kocqrz Ifwe want this to add up to Vp 0 we must have at rzrl Rs 6 If we start with a charge q a distance s from the center of a grounded conducting sphere Vp 0 we must place an image charge39 inside the sphere The image charge is opposite in sign to the charge outside and smaller in the ratio IVs and located at a distance Rzs from the center of the sphere For a sphere of radius R in the presence of a charge q but the sphere is not grounded we can place an additional image charge Q at the center of the sphere in addition to the charge qIUs to increase its potential from 0 to kQR We can replay this with in nite line charges using a line charge outside a conducting cylinder or two conducting parallel long cylinders See the next page We start with in nite line charges 7 and J as shown HE L I S lt s gt Now from an in nite line charge E 2k7tr and when we integrate lEdr we get V 2k 1nr C So at p we have Vp 2k ln r1 ln r2 2k7tlnr1rz constant When r1 r we are midway between the lines so V 0 and the constant is also zero We do the same deal as before letting ss R2 Just as before Eqs 1 5 we nd rzrl IVs 6 independent of the angle 9 Thus we have equipotential surfaces given by V 2k ln rlrz 2k lnsR 7 Eq 7 expresses the fact that there are a series of cylindrical equipotential surfaces between the two in nite lines of charge But expressing them in terms of the distances between two cylinders of radius R is algebraically a bit messy BS vector potential Ampere PH 316 MJM 1020 06 The generalized BiotSavart law is l B uo4nl d39 Jr X 0 u3 where l 1 Iquot r goes to the eld point r to the source pt We have shown before that 003 V lu 66 lN Xx 2 XX XX 2 32 so 1 can be written 2 B uo4nl d39 Jr X V l u Since V x fA E Vf x A f VxA we can rewrite 2 with A gt J and f gtlu as 3 Bu04nldtVxJuVxJu The 2nd term vanishes because J is a function of source coordinates and V acts on eld coordinates Thus 4 B v x FLO41 I d Jo The magnetic eld is derivable from the curl of a vector potential A B curl A and A FLO41 I d Jo In electrostatics we had a eld derivable from a scalar potential E grad V and v l41t80 ld pL There is a certain symmetry here with charges causing the scalar potential and the electric eld derived from the scalar potential and for steady currents currents cause the vector potential and the magnetic eld derivable from the vector potential Notice that between the two potentials so gt lu0 We can also show that curl B uo J by taking the curl of 4 and remembering that VXVXAVVOAVZA over 5 V x B VV0A uo4n ld c J V2 10 In 5 we have the freedom Grif ths pp 234 235 to set VoA 0 From p 50 at the bottom V2 10 41150 so 5 becomes curl B Ha I d 50 Jr Ho Jr This is ampere s law in differential form V x B uo Jr Ampere39s law How do you convert this to the integral form The integral form is lc Bod Ha I Ampere39s lavxi So Ampere39s law is derivable from the BS law and the BS law also implies VoB 0 This means the ux of B through any closed surface whatsoever must be zero It follows from B curl A and from the vector identity div curl V E 0 for any vector V


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