A) \[2{{E}_{0}}{{a}^{2}}\]
B) \[\sqrt{2}{{E}_{0}}{{a}^{2}}\]
C) \[{{E}_{0}}\,{{a}^{2}}\]
D) \[\frac{{{E}_{0}}{{a}^{2}}}{\sqrt{2}}\]
Correct Answer: C
Solution :
[c] Given \[\overrightarrow{E}={{E}_{0}}\,\,\hat{x}\] This shows that the electric field acts along \[+\,\,x\] direction and is a constant. The area vector makes an angle of \[45{}^\circ \] with the electric field. Therefore the electric flux through the shaded portion whose area is \[a\times \sqrt{2}a=\sqrt{2}\,\,{{a}^{2}}\] is \[\phi =\overrightarrow{E}.\overrightarrow{A}=EA\cos \theta ={{E}_{0}}(\sqrt{2}\,{{a}^{2}})\] \[\cos \,\,45{}^\circ ={{E}_{0}}(\sqrt{2}\,{{a}^{2}})\times \frac{1}{\sqrt{2}}={{E}_{0}}{{a}^{2}}\]You need to login to perform this action.
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