Question

In: Physics

In class, we learned that the electric field produced by a large, thin sheet of metal...

In class, we learned that the electric field produced by a large, thin sheet of metal with a uniform distribution of charge is E=k2(pi)(Q/A)R^    where an area A has total charge Q and R^ point away from the sheet.  Two very large, very thin sheets of metal are parallel to each other as shown below end-on. Both sheets are charged; sheet A has +1 nC for each 1m^2 of area and sheet B has -2nC for each 1m^2 of area Determine the electric field in the three regions I, II, III that is produced by all the charge.


Area I is on the left side of sheet A. Area II is in between sheet A and B. Area III is on the right side of sheet B.

Solutions

Expert Solution

Generl equation:

E = 2 k pi (Q/A) R^

----------------------------------

electric field due to the sheet A = E_A

electric field due to the sheet B = E_B

---------------------------------------------------------------------------------------------

Area I:

We must use the superposition:

E_I = (E_A) + (E_B)

==> E_I = 2 k pi ((Q_A)/A) (-R^) + 2 k pi ((Q_B)/A) (+R^)

(Note that (-R^) means that the direction of the E_A is negative and (+R^) means that the direction of the E_B is positive)

==> E_I = 2 k pi ((1e-9)/1) (-R^) + 2 k pi ((2e-9)/1) (+R^)

==> E_I = 2 * 8.99e9 * 3.1416 * ((1e-9)/1) (-R^) + 2 * 8.99e9 * 3.1416 * ((2e-9)/1) (+R^)

==> E_I = (-2 * 8.99e9 * 3.1416 * ((1e-9)/1) + 2 * 8.99e9 * 3.1416 * ((2e-9)/1)) R^

==> E_I = 56.5 R^ N/C

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Area II:

We must use the superposition:

E_II = (E_A) + (E_B)

==> E_II = 2 k pi ((Q_A)/A) (+R^) + 2 k pi ((Q_B)/A) (+R^)

==> E_II = 2 k pi ((1e-9)/1) (+R^) + 2 k pi ((2e-9)/1) (+R^)

==> E_II = 2 * 8.99e9 * 3.1416 * ((1e-9)/1) (+R^) + 2 * 8.99e9 * 3.1416 * ((2e-9)/1) (+R^)

==> E_II = (2 * 8.99e9 * 3.1416 * ((1e-9)/1) + 2 * 8.99e9 * 3.1416 * ((2e-9)/1)) R^

==> E_II = 169 R^ N/C

-----------------------------------------------------------------------------------------------

Area III:

We must use the superposition:

E_III = (E_A) + (E_B)

==> E_III = 2 k pi ((Q_A)/A) (+R^) + 2 k pi ((Q_B)/A) (-R^)

==> E_III = 2 k pi ((1e-9)/1) (+R^) + 2 k pi ((2e-9)/1) (-R^)

==> E_III = 2 * 8.99e9 * 3.1416 * ((1e-9)/1) (+R^) + 2 * 8.99e9 * 3.1416 * ((2e-9)/1) (-R^)

==> E_III = (2 * 8.99e9 * 3.1416 * ((1e-9)/1) - 2 * 8.99e9 * 3.1416 * ((2e-9)/1)) R^

==> E_III = -56.5 R^ N/C




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