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Chapter 21: Problem 16

The electric field of a flat sheet of charge is \(\sigma / 2 \epsilon_{0} .\) Yet the field of a flat conducting sheet-even a thin one, like a piece of aluminum foil- is \(\sigma / \epsilon_{0} .\) Explain this apparent discrepancy.

Short Answer

Expert verified
The difference in electric fields between a flat sheet of charge and a flat conducting sheet is because the conducting sheet allows the charges to move. These charges distribute themselves evenly over the two surfaces of the sheet and so, create an electric field on both sides. But for a non-conducting flat sheet, the charges are stationary on one surface, thus generating an electric field on one side only.

Step by step solution

01

Understand the properties of a conducting sheet

First, know that a conducting sheet (like aluminum) allows free movement of charge. If this sheet is charged, the charges will distribute evenly over the surface of the conductor. Because the charges move to the surfaces of the sheet, it effectively has two surfaces with charge density \(\sigma\). Hence, it creates an electric field both above and below the sheet.
02

Understand the properties of a flat sheet of charge

Conversely, a flat, non-conductive sheet of charge does not allow for charge movement. As such, it only has one 'effective' surface on which the charge resides, hence, only generate an electric field in one direction.
03

Explanation of the discrepancy

The flat conducting sheet creates an electric field of \(\sigma / \epsilon_{0}\) both above and below the sheet due to the movement of charges which settle on both surfaces of the sheet. On the other hand, the flat sheet of charge creates an electric field of \(\sigma / 2 \epsilon_{0}\) only on one side as the charges are stationary and do not move to the other side. This explains the discrepancy in the electric field.

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Most popular questions from this chapter

The electric field on the surface of a 10 -cm-diameter sphere is perpendicular to the sphere and has magnitude 47 kN/C. What's the electric flux through the sphere?

In Gauss's law, \(\oint \vec{E} \cdot d \vec{A}=q / \epsilon_{0},\) does the field \(\vec{E}\) necessarily arise only from charges within the closed surface?

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