P26_051.PDF

51. (a) We know from Eq. 26-7 that the magnitude of the electric ﬁeld is directlyproportional to the surface

charge density:

E = σ

ε 0 =

10 − 6 C / m 2

m 2 =1 . 7

10 6 V / m .

8 . 85

10 − 12 C 2 / N

Regarding the units, it is worth noting that a Volt is equivalent to a N

m/C.

(b) Eq. 26-23 yields

u = 1

2 ε 0 E 2 =13J / m 3 .

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 =13J / m 3

(d) The magnitude of force is

= dU

dx = uA .

(e) The force per unit area is

= u =13N / m 2 .

m, which explains how J/m 3

maybe set equal to N/m 2 in the above manipulation. We note, too, that the pressure unit N/m 2

is generallyknown as a Pascal (Pa).

(f) Combining our steps in parts (a) through (e), we have

= u = 1

2 ε 0 E 2

2 ε 0 σ

= σ 2

2 ε 0

6 . 0N / m 2

ε 0

which leads to σ = 2(8 . 85

10 − 12 )(6 . 0) = 1 . 0

10 − 5 C / m 2 .

Regarding units, it is worth noting that a Joule is equivalent to a N