A single isolated, large conducting plate has acharge per unit area σ on its surface. Because the plate is a conductor, the electric field at its surface is perpendicular to the surface and has magnitude E = σ/εo a.The field from a large, uniformly charged sheet with charge per unit area σ has magnitude E = σ/2εo. Why is there a difference? b.Regard the charge distribution on the conducting plate as two sheets of charge (one on each surface), each with charge per unit area σ. Find the electric field inside and outside the plate.
A single isolated, large conducting plate has acharge per unit area σ on its surface. Because the plate is a conductor, the electric field at its surface is perpendicular to the surface and has magnitude E = σ/εo a.The field from a large, uniformly charged sheet with charge per unit area σ has magnitude E = σ/2εo. Why is there a difference? b.Regard the charge distribution on the conducting plate as two sheets of charge (one on each surface), each with charge per unit area σ. Find the electric field inside and outside the plate.
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter25: Gauss’s Law
Section: Chapter Questions
Problem 47PQ: The infinite sheets in Figure P25.47 are both positively charged. The sheet on the left has a...
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A single isolated, large
a.The field from a large, uniformly charged sheet with charge per unit area σ has magnitude E = σ/2εo. Why is there a difference?
b.Regard the charge distribution on the conducting plate as two sheets of charge (one on each surface), each with charge per unit area σ. Find the electric field inside and outside the plate.
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