A) \[\frac{\text{o }\!\!|\!\!\text{ }}{\pi {{d}^{2}}}\]
B) \[\frac{\text{2o }\!\!|\!\!\text{ }}{\pi {{d}^{\,2}}}\]
C) \[\frac{4{{d}^{2}}}{\pi \text{o}{{\text{ }\!\!|\!\!\text{ }}^{\,2}}}\]
D) \[\frac{4\text{o }\!\!|\!\!\text{ }}{\pi {{d}^{2}}}\]
Correct Answer: D
Solution :
From Gausss theorem \[\int{E.ds=\text{o }\!\!|\!\!\text{ }}\] where E is electric field intensity, s the surface area, \[\text{o }\!\!|\!\!\text{ }\]the flux. Given, \[s=\pi {{r}^{2}}=\pi {{\left( \frac{d}{2} \right)}^{2}}=\frac{\pi {{d}^{2}}}{2}\] where r is radius and d the diameter. \[\text{o }\!\!|\!\!\text{ =}\,\text{E}\times \frac{\pi {{d}^{2}}}{4}\] \[\Rightarrow \] \[E=\frac{4\text{o }\!\!|\!\!\text{ }}{\pi {{d}^{2}}}\]
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