problem24_38

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 24.38: a) C = K0 A d gives us the area of the plates: Cd (5.00 10-12 farad)(1.50 10-3 m) = 8.475 10- 4 m 2 -12 2 2 K0 (1.00)(8.85 10 C / N m ) We also have C = K 0 A d = Q V , so Q = K 0 A(V d ). V d is the electric field A= between the plates, which is not to exceed 3.00 10 4 N C. Thus Q = (1.00)(8.85 10 -12 C 2 N m 2 )(8.475 10 -4 m 2 )(3.00 104 N C) = 2.25 10 -10 C b)Again, Q = K0 A(V d ) = 2.700 A(V d ). If we continue to think of V d as the electric field, only K has changed from part (a); thus Q in this case is (2.70)(2.25 10 -10 C) = 6.08 10 -10 C. ...
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problem24_38 - 24.38 a C = K0 A d gives us the area of the...

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