Ch23_p48 - is 8.0 × 10 7 N/C Therefore E shell E particle = E shell(2.4 3 2 E A = 8.0 × 10 7 N/C Using the value for E A noted above we find E

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48. Let E A designate the magnitude of the field at r = 2.4 cm. Thus E A = 2.0 × 10 7 N/C, and is totally due to the particle. Since E particle = q 4 πε o r 2 then the field due to the particle at any other point will relate to E A by a ratio of distances squared. Now, we note that at r = 3.0 cm the total contribution (from particle and shell)
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Unformatted text preview: is 8.0 × 10 7 N/C. Therefore, E shell + E particle = E shell + (2.4 / 3 ) 2 E A = 8.0 × 10 7 N/C Using the value for E A noted above, we find E shell = 6.6 × 10 7 N/C. Thus, with r = 0.030 m, we find the charge Q using E shell = Q 4 πε o r 2 ¡ Q = 6.6 × 10 − 6 C ....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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