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Question Two parallel conducting plates are separated by 1 mm and carry equal but opposite surface...

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Two parallel conducting plates are separated by 1 mm and carry equal but opposite surface charge densities. If the potential difference between them is 5 V, what is the magnitude of the surface charge density on each plate?

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Answer #1

The potential difference between the two plates=V =Ed

V=E*(0.001)

5/0.001=E

5000V/m=E

Magnitude of E =Surface charge density / absolute permittivity

absolute permittivity =8.85*10^-12

E*absolute permittivity =Surface charge density

5000*8.85*10^-12=Surface charge density

44250*10^-12C/m^2 =Surface charge density

Surface charge density=44.3*10^-9 C/m^2

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