Question

gaussian surface of uniformly charged infinite plane sheet ​

Answer 1

Explanation:

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER}\mid}}}$

$Electric \: \: field \: \: due \: \: to \: \: uniformly \: \: \\ charged \: \: infinite \: \: plane \: \: sheet. \\ Suppose \: \: a \: \: thin \: \: non-conducting \: \: \\ infinite \: \: sheet \: \: of \: \: uniform \: \: \\ surface, \: \: charge \: \: density \: \: \sigma$

$Electric \: \: field \: \: intensity \: \: on \: \: either \: \: \\ side \: \: of \: \: the \: \: sheet \: must \: \: be \: \: perpendicular \\ to \: \: the \: \: plane \: \: of \: \: sheet \: \: having \: \: \\ same \: \: magnitude \: \: of \: \: all \: \: points \: \: \\ equidistant \: \: from \: \: the \: \: sheet. \: \: \\ Let \: \: P \: \: be \: \: any \: \: point \: \: at \: \: distance \: \: 'r' \\ from \: \: the \: \: sheet. \: \: Let \: \: the \: \: small \: \: \\ area \: \: element \: \: be \: \: \vec{E} = ds \widehat{n} .$

$Where \: \: \vec{E} \: \: and \: \: \widehat{n} \: \: are perpendicular, \\ on \: \: the \: \: surface \: \: of \: \: imagined \: \: cylinder, \\ so \: \: electric \: \: flux \: \: is \: \: zero. \\ Also, \: \: \vec{E} \: \: and \: \: \widehat{n} \: \: are \: \: parallel \: \: \\ on \: \: the \: \: two \: \: cylindrical \: \: edges \: \: P \: \: and \: \: Q, \: \\ Which \: \: contributes \: \: Electric \: \: flux.$

$\therefore \: Electric \: \: flux \: \: over \: \: the \: \: edges \\ P \: \: and \: \: Q \: \: of \: \: the \: \: cylinder \: \: is : \\ \\ 2 \phi = \frac{q}{ \epsilon_{0}} \implies 2 \oint \vec{E} \vec{ds \: \: \: } = \frac{q}{ \epsilon_{0}}$

$2 \oint \vec{E}. \vec{ds} = \frac{q}{ \epsilon_{0}} \\ \\ (\vec{E} \perp \vec{ds}) \\ \\ 2 \vec{E}\pi {r}^{2} = \frac{q}{ \epsilon_{0}} \\ \\ \boxed{ \vec{E} = \frac{q}{2\pi {r}^{2} \epsilon_{0}} }$

$The \: \: charge \: \: density \: \: \sigma = \frac{q}{S} \\ \\ where \: \: S \: \: = \: \: area \: \: of \: \: circle \\ \\ {E} = \frac{\pi {r}^{2} \sigma}{2\pi \epsilon_{0} {r}^{2} } = \frac{ \sigma}{2 \epsilon_{0}} \\ \\ \vec{E} = \frac{ \sigma}{2 \epsilon_{0}} \widehat{n} \\ \\ where \: \: \widehat{n} \: \: is \: \: unit \: \: vector \: \: normal \: \: \\ to \: \: plane \: \: and \: \: going \: \: away \: \: from \: \: it$

$When \: \: \sigma > 0 , \: \: \vec{E} \: \: is \: \: directed \: \: \\ away \: \: from \: \: both \: \: the \: \: sides, \\ Hence \: \: electric \: \: field \: \: intensity \: \: \\ is \: \: independent \: \: of \: \: r . \\ Note :- For \: \: conducting \: \: sheet, \: \: \\ the \: \: surface \: \: charge \: \: density \: \: on \: \: \\ both \: \: the \: \: surface \: \: willm \: be \: \: same \: \: \\ \\ \huge\boxed{ \therefore \vec{E} = \frac{\sigma}{\epsilon_{0}} }$