Exercise 22.30 Part A Two very large, nonconducting plastic sheets, each 10.0 cm

Question

Exercise 22.30 Part A Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities 1, 2, 3 and on their surfaces, as shown in the following figure(Figure 1). These surface charge densities have the values 1 =-6.70 pC/m2 , 2 = 5.00 pC/m2 , 7; = 1.30 C/m2 , and 4 4.00 C/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 What is the magnitude of the electric field at point A, 5.00 cm from the left face of the left-hand sheet? Express your answer to three significant figures and include the appropriate units E-Value Units Figure 1 of 1 Submit My Answers Give Up Part B What is the direction of the electric field at point A, 5.00 cm from the left face of the left-hand sheet? to the left. to the right. upwards downwards 10 cm 12 cm 10 cm Submit My Answers Give Up

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 = q/(A*0) q/A = so E = /0
so the answer to a) is

(1 + 2 + 3 + 4)/0 =
3.6*10^-6/0 N/C

Towards right

For 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

3.6*10^-6/0 + E = (1 + 2)/0

E = (3.6*10^-6 +6.70*10^-6 - 5.00*10^-6)/0

=5.3*10^-6/ 0
Towards left

In the middle of the right hand sheet, just add 3 to the right hand side (charge enclosed)
3.6*10^-6/0 + E = (1 + 2 + .3)/0

E = -4.0*10^-6/ 0

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