# hw2.7 - HW2.7 Given Find a 2cm E x 12 2 0 E x 7 1 4C m 2 E...

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Unformatted text preview: HW2.7 Given: Find: a 2cm E x @ 12 ? 2 0 E x @ 7 ? 1 4C / m 2 E x @ 5 ? p 8C / m 2 E x @ 3 ? [email protected] ? [email protected] ? a ? b ? c ? d ? Solutions a.) To find E x @ 12 , E x @ 7 , E x @ 5 , E x @ 3 E x @ 3 & E x @ 7 we can use the following formula: E 2 0 E x @ 12 for all parts p 2 0 E x @ 7 E x @ 5 0 p 2 0 p 1 1 2 0 2 0 1 p 1 2 0 2 0 (inside the conductor) E x @ 12 E x @ 7 p 1 2 0 p 1 2 0 E x @ 3 p 2 0 b.) To E x @ 3 p 1 2 0 (inside the conductor) [email protected] 0 [email protected] 1 p 1 2 0 2 0 p 2 0 1 p 1 2 0 2 0 [email protected] p 1 2 0 find a , b , c & d we can use the following formula: EdA a Q 0 aA 0 p 1 A 2 Q EdA 0 b 1 a EA cA 0 p 1 2 c b 2 d 2 c a A 2 0 0 note: minuses are there for direction of E field p 1 a b 1 c EA p 1 2 0 A c A 0 ...
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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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hw2.7 - HW2.7 Given Find a 2cm E x 12 2 0 E x 7 1 4C m 2 E...

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