hw2.5 - HW2.5 Given: Find: R1 0.6m R2 2.1m E0.9 ? 1.45C / m...

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HW2.5 Given: Find: m R 6 . 0 1 ? 3 . 0 E m R 1 . 2 2 ? 9 . 0 E 3 / 45 . 1 m C  ? 2 . 4 E Solutions a.) To find , & we can use the following formula: 3 . 0 E 9 . 0 E 2 . 4 E 0 enc Q EdA     r R r L r LR Lr A V A Q E enc 0 2 1 2 0 2 1 2 0 0 3 . 0 2 2     0 2 0 2 1 2 3 . 0 r R r E since 0 1 R r 0 3 . 0 E   r R r E 0 2 1 2 9 . 0 2   r R R E 0 2 1 2 2 9 . 0 2 it’s not r because r is outside the cylinder i.e. 2 R 0 outside
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Part 2 b.) To find & when the hole is at (x, y) = (0, 0.9) m and the point is (x, y, z) = (0, 4.2, 0) m we can use the following formula: x E y E 0 cause E is straight up x E 0 enc Q EdA First let’s find E if there was no hole: 0 2 2 0 0 2 r R A V A Q E enc Now let’s find E that hole took out:  0 2 1 0 0 2 h enc r r R A V A Q E  h h y r r R r R r r R r R E 2 1 2 2 0 0 2 1 0 2 2 2 2 2 h y r r R r R E 2 1 2 2 0 2 Note: r = y position of the point = y position of the hole h r
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Part c.) To find
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This note was uploaded on 01/26/2012 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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hw2.5 - HW2.5 Given: Find: R1 0.6m R2 2.1m E0.9 ? 1.45C / m...

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