Two large, nonconducting plates are suspended8.51 cm apart. Plate 1 has an area charge density of +86.8 µC/m2, and plate 2 has an area charge density of +10.1 µC/m2. Treat each plate Plate 1 Plate 2 as an infinite sheet. How much electrostatic energy UF is stored in 3.79 cm' of the space in region A? Region A Region B Region C UE = J What volume V of the space in region B stores an equal amount of energy? V = cm3
![Two large, nonconducting plates are suspended 8.51 cm apart.
Plate 1 has an area charge density of +86.8 µC/m², and plate
Plate 1
Plate 2
2 has an area charge density of +10.1 µC/m2. Treat each plate
as an infinite sheet.
How much electrostatic energy UF is stored in 3.79 cm of the
space in region A?
Region A
Region B
Region C
UF =
J
%3D
What volume V of the space in region B stores an equal
amount of energy?
V =
cm3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd33fc653-3fa9-47ed-89c8-625102275194%2Fbfad015f-8922-4101-bceb-74482ce079c5%2F3ay5oqp_processed.jpeg&w=3840&q=75)
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given :
areal charge density of plate 1 , 1 = 86.8 C/m2
areal charge density of plate 2 , 2 = 10.1 C/m2
distance between the plates d = 8.51 cm = 8.51 x10-2 m
both of the sheets are treated as infinite sheets .
the Electric field produced by such an infinite sheet E =
where is the areal surface charge density
= permittivity of the free space = 8.85 x 10-12farad/meter
let the electric field due to plate A and B be E1 and E2 respectively.
E1 = = = 4.904 x 106 N/C
E2 = = =0.571 x 106 N/C
net electric field = E = E1 +E2 =4.904 x 106 + 0.571 x 106 (N/C) = 5.475 x 10 6 N/C
charge density in region A= UA= = = 132.642 J/m3
amount of energy stored in 3.79 cm3 U3.79 = 132.642 x 3.79 x10-6 = 5.027 x 10-4 J
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