ch25-p052

ch25-p052 - 52(a The electric field E1 in the free space...

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Unformatted text preview: 52. (a) The electric field E1 in the free space between the two plates is E1 = q/ε0A while that inside the slab is E2 = E1/κ = q/κε0A. Thus, bg V0 = E1 d − b + E2b = FG q IJ FG d − b + b IJ , H ε AK H κ K 0 and the capacitance is (8.85×10−12 C2 /N ⋅ m2 ) (115×10−4 m2 ) ( 2.61) = 13.4pF. ε 0 Aκ q C= = = V0 κ ( d − b ) + b ( 2.61)( 0.0124m − 0.00780m ) + ( 0.00780m ) (b) q = CV = (13.4 × 10–12 F)(85.5 V) = 1.15 nC. (c) The magnitude of the electric field in the gap is E1 = 115 × 10−9 C . q = = 113 × 104 N C . . −12 C 2 ε 0 A 8.85 × 10 N⋅m2 115 × 10−4 m2 d ic h (d) Using Eq. 25-34, we obtain E2 = E1 κ = 113 × 104 N C . = 4.33 × 103 N C . 2.61 ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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