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Question
What is the electric field just outside a single conducting plate with surface charge density \[\sigma\]?
Options
\[E = \frac{\sigma A}{\varepsilon_0}\]
\[E = \frac{\sigma}{\varepsilon_0}\]
\[E = \frac{2\sigma}{\varepsilon_0}\]
\[E = \frac{\sigma}{2\varepsilon_0}\]
MCQ
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Solution
Only one face of the pillbox carries flux because the field is zero inside the conductor. Gauss's Law gives \[EA = \sigma A/\varepsilon_0\], hence \[E = \sigma/\varepsilon_0\].
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