Consider a very thin rectangular conducting plate whose length is 109.0 cm, width is 86.0 cm, and has a total charge of −58.0 nC. The plate lies in the horizontal plane (xz plane) and has a uniform charge density. Find the electric field vector at points just above the plate and just below the plate. Express your answer in vector form.

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aristocles aristocles · Answer 1 · 2019-10-09T02:22:34+02:00

Answer:

electric field just above the plate is given as

[tex]E = -3495.6 \hat k[/tex]

Now electric field just below the plate is given as

[tex]E = 3495.6 \hat k[/tex]

Explanation:

Charge density of the plates is given as

[tex]\sigma = \frac{Q}{A}[/tex]

here we know that

[tex]Q = -58 nC[/tex]

Area = (length)(width)

[tex]A = 1.09\times 0.86 m^2[/tex]

[tex]A = 0.9374 m^2[/tex]

Now we have

[tex]\sigma = \frac{-58 nC}{0.9374}[/tex]

[tex]\sigma = -61.87 nC/m^2[/tex]

now electric field due to a thin sheet is given by

[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]

[tex]E = \frac{61.87\times 10^{-9}}{2(8.85 \times 10^{-12}}[/tex]

[tex]E = 3495.6 N/C[/tex]

Now electric field just above the plate is given as

[tex]E = -3495.6 \hat k[/tex]

Now electric field just below the plate is given as

[tex]E = 3495.6 \hat k[/tex]