If the radius of the Gaussian …
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If the radius of the Gaussian surface enclosing a charge is halved , how does the electric flux through the Gaussian surface charge
Posted by Alisu Jingru 6 years, 4 months ago
- 2 answers
Amisha Shrivastava 6 years, 4 months ago
Electric Flux is independent or size and shape of the gaussian surface. It depends on the magnitude of charge enclosed within the surface.
Therefire,on changing radius electric flux will remain same.
Therefire,on changing radius electric flux will remain same.
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Sia ? 6 years, 4 months ago
Electric flux {tex}{\phi _E}{/tex} is given by
{tex}{\phi _E} = \oint {\overrightarrow E } \cdot \overrightarrow {ds} = \frac{Q}{{{\varepsilon _0}}}{/tex}
where {tex}\theta{/tex} is total charge inside the closed surface.
{tex}\therefore{/tex} On changing the radius of sphere, the electric flux through the Gaussian surface remains same.
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