Question

2 02 = 5.31 μο/m 2, and The figure shows, in cross section, three infinitely large nonconducting sheets on which Charge is uniformly spread. The surface charge densities are σι = 2.10 μα -4.87 μC/m2, and distance L-1.08 cm. what are the (a) x and (b) y components of the net electric field at point = L/2 os 2L I.

Answer 1

The electric field due to infinitely large non-conducting sheet of charge density $\dpi{100} \sigma$ at any point at distance r is given by

$\dpi{100} E = \frac{\sigma}{2 \epsilon _{0}}$

And the electric field will be perpendicular to the surface.

Where, $\dpi{100} \epsilon _{0} = 8.85 \times 10^{-12} \, \, farad/m$ is permittivity of free space .

so,

the electric field due to first sheet at P will be

$\dpi{100} E_{1} = \frac{\sigma_{1}}{2 \epsilon _{0}} = \frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 1.186 \times 10^{5} \, \, N/C$

the electric field due to second sheet at P will be

$\dpi{100} E_{2} = \frac{\sigma_{2}}{2 \epsilon _{0}} = \frac{5.31 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 3 \times 10^{5} \, \, N/C$

the electric field due to third sheet at P will be

$\dpi{100} E_{3} = \frac{\sigma_{3}}{2 \epsilon _{0}} = \frac{-4.87 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} =- 2.75 \times 10^{5} \, \, N/C$

And direction of all the three electric field will be along the y - axis.

So,   

  $\dpi{100} E_{net} = E_{1} + E_{2} + E_{3} = (1.186 + 3-2.75) \times 10^{5} = 1.436 \times 10^{5} \, \, N/C$

In the positive y - direction.

(a) The x - component of the field is zero.

(b) The y - component of the field is 1.436 x 10⁵ N/C.

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