Question 22.30 Gauss

# Question 22.30 Gauss - σ = E 1 4 3 2 2 2 2 2 = A E At...

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Two very large, non conducting plastic sheets, each 10.0 cm thick, carrying uniform charge densities σ 1 σ 2 σ 3 σ 4 , on their surfaces, as shown in the figure. These charge densities have the value of: σ 1 = -6.00μC/m 2 σ 2 = +5.00 μC/m 2 σ 3 = +2.00 μC/m 2 σ 4 = +4.00 μC/m 2 Use Gauss’s Law to find the magnitude and direction of the electric field at the following points far from the edges of these charge sheets (to prevent fringing): A) 5.00cm from the left face of the left-hand sheet B) 1.25cm from the inner surface of the right hand sheet C) Point C , in the middle of the right hand sheet. Q 22.30

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The net electric field is the vector sum of the fields due to each of the four sheets of charge . The electric field of a large sheet of charge is: The field is directed away from a positive sheet and toward a negative sheet. 0 2 ε
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Unformatted text preview: σ = E 1 4 3 2 2 2 2 2-+ + = A E At point A the electric field is: [ ] C N x m C m C m C m C E A / 10 82 . 2 / 6 / 4 / 2 / 5 2 1 5 2 2 2 2 =-+ + = μ This field points to the left. . σ 1 = -6.00μC/m 2 σ 2 = +5.00 μC/m 2 σ 3 = +2.00 μC/m 2 σ 4 = +4.00 μC/m 2 2 4 3 1 2 2 2 2 ε σ-+ + = B E [ ] C N x m C m C m C m C E B / 10 95 . 3 / 5 / 4 / 2 / 6 2 1 5 2 2 2 2 =-+ + = μ At point B At point C 3 2 1 4 2 2 2 2--+ = c E [ ] C N x m C m C m C m C E c / 10 69 . 1 / 2 / 5 / 6 / 4 2 1 5 2 2 2 2 =--+ = σ 1 = -6.00μC/m 2 σ 2 = +5.00 μC/m 2 σ 3 = +2.00 μC/m 2 σ 4 = +4.00 μC/m 2 Points to the left Points to the left...
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Question 22.30 Gauss - σ = E 1 4 3 2 2 2 2 2 = A E At...

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