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7/10 points v Previous Answers OSUNIPHYS1 23.3.P.052. My Notes A long copper cylindrical shell of inner radius 4 cm and outer

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Answer #1

(a)

E = 0

---------------------------------------------

(b)

at distance of 2.5 cm

q = \lambda L

so,

E (2\pirL) = \lambda L / \epsilon o

E = \lambda / 2\pir\epsilono

so,

E = 3e-12 / 2\pi * 0.025 * 8.85e-12

E = 2.158 N/C

--------------------------------------------

(c)

E = 0 N/C

--------------------------------------------

(d)

for r = 13 cm

E = \lambda / 2\pir\epsilono

so,

E = 3e-12 / 2\pi * 0.13 * 8.85e-12

E = 0.415 N/C

--------------------------------------------------------

(e)

at r = 22 cm

E = \lambda / 2\pir\epsilono

so,

E = 3e-12 / 2\pi * 0.22 * 8.85e-12

E = 0.245 N/C

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