Two very large non-conducting plastic sheets, each 10.0 cm thick, carry uniform

Question

Two very large non-conducting plastic sheets, each 10.0 cm thick, carry uniform charge densities sigma 1, sigma 2, sigma 3, sigma 4 on their surfaces, as shown below. The surface charge densities have the values sigma 1=-6.00 mu C/m2,sigma 2 = + 5.00 mu C/m2, sigma 3 = + 2.00 mu C/m2, sigma 4 = + 4.00 mu C/m2. Find the magnitude and direction of the electric field at the following points, far from the edges of these sheets. Point A, 5.00 cm from the left face of the left-hand sheet. Point B, 1.25 cm from the inner surface of the right-hand sheet. 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 E*A = q/e0 = q/(A*e0) q/A = s so E = s/e0 so the answer to a) is (s1 + s2 + s3 + s4)/e0 = 5*10^-6/e0 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/e0 + E = (s1 + s2)/e0 E = (-5*10^-6 - 6.00*10^-6 + 5.00*10^-6)/e0 In the middle of the right hand sheet, just add s3 to the right hand side (charge enclosed) 5*10^-6/e0 + E = (s1 + s2 + s3)/e0

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