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

and on their surfaces, as

shown in Fig. E22.32. These surface

charge densities have the values

s1= -6.00 mC>m2,

s2 = +5.00 mC>m2,

s3 =+2.00 mC>m2

and s4 = +4.00

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 5.00 cm from the left face of

the left-hand sheet; (b) point 1.25 cm from the inner surface of the

right-hand sheet; (c) point in the middle of the right-hand sheet.

Explanation / Answer

The two very large non conducting plastic sheets carries uniform charge densities

S1= -6.00 microC/m2,     S2= +5.00 microC/m2 , S3= +2.00 microC/m2 ,       S4= +4.00 microC/m2

eps0= 8.854x10-12

Gauss's law, integral over closed surface E. da = q/ eps0,           EA= q / eps0,

           E = q / Aeps0 = S / eps0

a) The electri field at a point A, 5.00 cm from the left of left hand sheet.

     Let we compress two sheets into a single sheet and to get uniform charge density,

              S= -S1 - S2 - S3 -S4

                  = +6.00 - 5.00 - 2.00 - 4.00

                      = -5.00 microC/m2

    the magnitude and direction of Electric field at A,    E = S / eps0

                             E = -5.00x10-6 / 8.854x10-12

                                               E = -5.64x105 N/C pointing left towards A.

b) Electric field at the point B, 1.25 cm from the inner surface of the right hand sheet,

            Snet1 = S1 + S2 = -6.00 + 5.00 = -1.00 microC/m2

            Snet2 = -S3 - S4 = -4.00 - 2.00 = -6.00 microC/m2

               E= (Snet1 + Snet2)/2eps0

                     = (-1.00 - 6.00)x10-6 / 2x8.854x10-12

                     = -3.95x105 N/C   pointing to the left of right sheet.

c) The electric field at the point C, in middle of the right sheet,

           compress left sheet with right sheet, then

          Snet1 = S1 + S2+ S3 = -6.00 + 5.00 + 2.00 = +1.00x10-6 C/m2

          Snet2 = -S4

            E= (Snet1 + Snet2)/2eps0 = (+1.00 - 4.00)x10-6/ 2x8.854x10-12

                  E= -1.69x105 N/C   pointing towards left.

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