G Two very large, nonco x C Two very Large, Nonco x er Dashboard C www.chegg.co

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G Two very large, nonco x C Two very Large, Nonco x er Dashboard C www.chegg.co 22-problem-30E-5olution-9780133969290 m/homework-help/University-Physics-14th-edition-chapter Apps G3 A list of all the bullshit The Universal Law ot Bookmark Show all steps: DON :E Chapter 22, Problem 30E Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities s1, s2, s3, and s4 on their surfaces (Fig. E22.30). These surface charge densities have the values 01 6.00 HC/m2, o2 +5.00 HC/m2, a3 2.00 HC/m2, and o4 +4.00 HC/m2. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets: (a) point A, 5.00 cm from the left face of the left-hand sheet (b) point B 25 cm from the inner surface of the right-hand sheet; (c) point C, in the middle of the right-hand sheet. Figure E22.30 0 cm 12 cm 10 cm Type here to search 6:46 PM

Explanation / Answer

by symmetry, the field is uniform with respect to location parallel to the sheets and is normal to the sheet surface. so E*dA = E*A; this equals the total charge enclosed in a volume of area A

E*A = q/0 Or. E= q/(A*0)

q/A = so

E = /0

a)

E = (1 + 2 + 3 + 4)/0 = 5*10-6/0 N/C

b)

the point is between the sheets. Place the gaussian surface so that the one end is outside the sheets and the other at the specified location between the sheets. Since we know the field on the outside surface the surface integral of field is

5*10-6/0 + E = (1 + 2)/0
E = (-5*10-6 - 6.00*10-6 + 5.00*10-6)/0

C)

In the middle of the right hand sheet, just add 3 to the right hand side (charge enclosed)
5*10-6/0 + E = (1 + 2 + 3)/0 = 1*10-6

E= -4*10-6 NN/C

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