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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.

asked
User Shawyeok
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7.5k points

1 Answer

6 votes

Answer:

electric field just above the plate is given as


E = -3495.6 \hat k

Now electric field just below the plate is given as


E = 3495.6 \hat k

Step-by-step explanation:

Charge density of the plates is given as


\sigma = (Q)/(A)

here we know that


Q = -58 nC

Area = (length)(width)


A = 1.09* 0.86 m^2


A = 0.9374 m^2

Now we have


\sigma = (-58 nC)/(0.9374)


\sigma = -61.87 nC/m^2

now electric field due to a thin sheet is given by


E = (\sigma)/(2\epsilon_0)


E = (61.87* 10^(-9))/(2(8.85 * 10^(-12))


E = 3495.6 N/C

Now electric field just above the plate is given as


E = -3495.6 \hat k

Now electric field just below the plate is given as


E = 3495.6 \hat k

answered
User Vmassuchetto
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8.2k points