ch24-p093

# ch24-p093 - 93. (a) For r > r2 the field is like that of a...

This preview shows pages 1–2. Sign up to view the full content.

(b) To find the potential in the region r 1 < r < r 2 , first use Gauss’s law to find an expression for the electric field, then integrate along a radial path from r 2 to r . The Gaussian surface is a sphere of radius r , concentric with the shell. The field is radial and therefore normal to the surface. Its magnitude is uniform over the surface, so the flux through the surface is Φ = 4 π r 2 E . The volume of the shell is 43 2 3 1 3 π bg c h rr , so the charge density is ρ = 3 4 2 3 1 3 Q π c h , and the charge enclosed by the Gaussian surface is qr r Q = F H G I K J −= F H G I K J 4 3 3 1 3 3 1 3 2 3 1 3 π c h . Gauss’ law yields () 33 2 11 0 23 3 21 0 4. 4 Q rE Q E rr r ε ⎛⎞ −− π= = ⎜⎟ −π ⎝⎠ If V s is the electric potential at the outer surface of the shell ( r = r 2 ) then the potential a distance r from the center is given by VV E d rV Q r r r dr V Q rrr r r r ss r r r r s =− F H G I K J −+− F H G I K J z z 4 1 4 1 22 02 3 1 3 1 3 2 3 1 3 2 2 2 1 3 1 3 2 π π . The potential at the outer surface is found by placing r = r 2 in the expression found in part (a). It is V s = Q /4 πε 0 r 2 . We make this substitution and collect terms to find

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

### Page1 / 2

ch24-p093 - 93. (a) For r > r2 the field is like that of a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online