In: Physics
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.
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
-----------------------------------------------------------------------------------------------
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