Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform

Question

Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities ?1,?2,?3 and ?4 on their surfaces, as shown in the following figure(Figure 1) . These surface charge densities have the values ?1 = -6.60?C/m2 , ?2=5.00?C/m2, ?3 = 3.30?C/m2 , and ?4=4.00?C/m2. Find the magnitude and direction of the electric field at the following points, far from the edges of these sheets. (Use superposition, rather than Gauss's Law, for this problem.)

What is the magnitude of the electric field at point A, 5.00 cm from the left face of the left-hand sheet?

What is the magnitude of the electric field at point B, 1.25 cm from the inner surface of the right-hand sheet?

What is the magnitude of the electric field at point C, in the middle of the right-hand sheet?

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

a) E*A = q/?0 = q/(A*?0) q/A = ? so E = ?/?0 so the answer to a) is (?1 + ?2 + ?3 + ?4)/?0 = 5*10^-6/?0 N/C=5.17*10^-19 N/C

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

5*10^-6/?0 + E = (?1 + ?2)/?0
E = (-5*10^-6 - 6.00*10^-6 + 5.00*10^-6)/e0=-6.787x10^-19 N/C

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

=>E=

(?1 + ?2 + ?3)/? - 5*10^-6/?0

=(-6+5+2-5)x10^-6/eo

=-4.52x10^-19 N/C

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