P26_066 - 66. (a) Since the field is constant and the...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 66. (a) Since the field is constant and the capacitors are in parallel (each with 600 V across them) with identical distances (d = 0.00300 m) between the plates, then the field in A is equal to the field in B: V E= = 2.00 × 105 V/m . d (b) See the note in part (a). (c) For the air-filled capacitor, Eq. 26-4 leads to σ= q 2 = ε0 E = 1.77 × 10−6 C/m . A (d) For the dielectric-filled capacitor, we use Eq. 26-27: σ = κε0 E = 4.60 × 10−6 C/m . 2 (e) Although the discussion in the textbook (§26-8) is in terms of the charge being held fixed (while a dielectric is inserted), it is readily adapted to this situation (where comparison is made of two capacitors which have the same voltage and are identical except for the fact that one has a dielectric). The fact that capacitor B has a relatively large charge but only produces the field that A produces (with its smaller charge) is in line with the point being made (in the text) with Eq. 26-32 and in the material that follows. Adapting Eq. 26-33 to this problem, we see that the difference in charge densities between parts (c) and (d) is due, in part, to the (negative) layer of charge at the top surface of the dielectric; consequently, σ = 1.77 × 10−6 − 4.60 × 10−6 = −2.83 × 10−6 C/m . 2 ...
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