Unformatted text preview: ive. For full credit, you must clearly show (once) how Gauss’s Law is used to justify
any equations used.
(a) [5pt] Find E at
Since 3.5 cm. is inside the inner cylinder, and no charge is inside, (b) [5pt] Find E at R = 7.0 cm.
R Apply Gauss’s Law to a cylinder of radius
. The electric
flux is through the sides is
with
. The charge
inside is
with
, and Gauss’s Law gives
, so R1 (c) [5pt] Find E at R = 14.0 cm.
The same equation applies, but with net charge now
The radius is now twice what it was before
, with in the previous a...
View
Full Document
 Spring '12
 ScottYost
 Physics, Charge, Energy, Electric charge, 1.0 cm, 5.0 m, 6pt, 3.5 cm

Click to edit the document details