This preview shows page 1. Sign up to view the full content.
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 airﬁlled capacitor, Eq. 264 leads to
σ= q
2
= ε0 E = 1.77 × 10−6 C/m .
A (d) For the dielectricﬁlled capacitor, we use Eq. 2627:
σ = κε0 E = 4.60 × 10−6 C/m .
2 (e) Although the discussion in the textbook (§268) 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. 2632 and in
the material that follows. Adapting Eq. 2633 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.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details