Assignment1_Solution-ch3

Assignment1_Solution-ch3 - now turn z into r. .. 3.28 3.35...

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3.21 Every point on the ring is the same distance from a point on the z-axis, so the potential at z is Electric field along z-axis: By symmetry, it's only in the z- direction 3.22 By spherical symmetry all axes are equivalent, so we can use our result above for a single ring on the z-axis, and integrate over elemental rings. Notice that while the ring in 3.21 was at z=0, now we're integrating over rings at a height z=-R to z=R : Now, we need to make sure that we evaluate this correctly for z<R and z>r: Assignment 1 Solution Wednesday, October 01, 2008 9:11 PM Assignment solutions Page 1
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(Because, again, this problem is spherically symmetric, we can
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Unformatted text preview: now turn z into r. .. 3.28 3.35 Here, it's just the sum of the three line integrals over the line segments: Assignment solutions Page 2 3.43 This goes as 1/r in the limit r->0, so we know that it behaves like a point charge---i.e. a delta function of charge density---at the origin. Assignment solutions Page 3 V is the product of two scalar functions, one of which is 1/r. But we don't have an identicy for the Laplacian of (fg) on the inside cover of the book. What we do have is On the RHS we have the divergence of a scalar times a vector function. We have an identity for that: Using this, with Assignment solutions Page 4...
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Assignment1_Solution-ch3 - now turn z into r. .. 3.28 3.35...

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