| The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field E is applied at an angle\[\frac{\pi }{4}\]with the vertical direction. The electric flux through the curved surface of the hemisphere is [JEE ONLINE 19-05-2012] |
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A) \[\pi {{a}^{2}}E\]
B) \[\frac{\pi {{a}^{2}}E}{\sqrt{2}}\]
C) \[\frac{\pi {{a}^{2}}E}{2\sqrt{2}}\]
D) \[\frac{\left( \pi +2 \right)\pi {{a}^{2}}E}{{{\left( 2\sqrt{2} \right)}^{2}}}\]
Correct Answer: B
Solution :
| [b] We know that,\[\phi =\oint{E.dS=E}\oint{dS}\cos {{45}^{o}}\] |
| In case of hemisphere\[{{\phi }_{curved}}={{\phi }_{circular}}\] |
| Therefore, \[{{\phi }_{curved}}=E\pi {{a}^{2}}\frac{1}{\sqrt{2}}=\frac{E\pi {{a}^{2}}}{\sqrt{2}}\] |
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