P26_051 - 51. (a) We know from Eq. 26-7 that the magnitude...

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Unformatted text preview: 51. (a) We know from Eq. 26-7 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. 26-23 yields 1 ε0 E 2 = 13 J/m3 . 2 u= (c) The energy U is the energy-per-unit-volume 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.

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