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Unformatted text preview: 51. (a) We know from Eq. 267 that the magnitude of the electric ﬁeld is directly proportional to the surface
charge density:
σ
15 × 10−6 C/m2
E=
=
= 1.7 × 106 V/m .
ε0
8.85 × 10−12 C2 /N · m2
Regarding the units, it is worth noting that a Volt is equivalent to a N·m/C.
(b) Eq. 2623 yields
1
ε0 E 2 = 13 J/m3 .
2 u= (c) The energy U is the energyperunitvolume multiplied by the (variable) volume of the region
between the layers of plastic food wrap. Since the distance between the layers is x, and we use A
for the area over which the (say, positive) charge is spread, then that volume is Ax. Thus,
U = uAx where u = 13 J/m3 . (d) The magnitude of force is
F= dU
= uA .
dx (e) The force per unit area is
F
A = u = 13 N/m2 . Regarding units, it is worth noting that a Joule is equivalent to a N·m, which explains how J/m3
may be set equal to N/m2 in the above manipulation. We note, too, that the pressure unit N/m2
is generally known as a Pascal (Pa).
(f) Combining our steps in parts (a) through (e), we have
F
A
6.0 N /m2
which leads to σ = 1
ε0 E 2
2
2
1
σ2
σ
=
ε0
2
ε0
2 ε0 = u=
= 2(8.85 × 10−12 )(6.0) = 1.0 × 10−5 C/m2 . ...
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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, Charge

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