P26_044

# P26_044 - 44(a The electric ﬁeld E1 in the free space...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 44. (a) The electric ﬁeld E1 in the free space between the two plates is E1 = q/ε0 A while that inside the slab is E2 = E1 /κ = q/κε0 A. Thus, V0 = E1 (d − b) + E2 b = q ε0 A d−b+ b κ , and the capacitance is C = = = q ε0 Aκ = V0 κ(d − b) + b 8.85 × 10−12 C2 N·m2 (115 × 10−4 m2 )(2.61) (2.61)(0.0124 m − 0.00780 m) + (0.00780 m) 13.4 pF . (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 = q 1.15 × 10−9 C = = 1.13 × 104 N/C . 2 ε0 A 8.85 × 10−12 NC 2 (115 × 10−4 m2 ) ·m (d) Using Eq. 26-32, we obtain E2 = E1 1.13 × 104 N/C = = 4.33 × 103 N/C . κ 2.61 ...
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online