A) \[\frac{\phi }{\pi {{d}^{2}}}\]
B) \[\frac{2\phi }{\pi {{d}^{2}}}\]
C) \[\frac{4{{d}^{2}}}{\pi {{\phi }^{2}}}\]
D) \[\frac{4\phi }{\pi {{d}^{2}}}\]
Correct Answer: D
Solution :
From Gauss's theorem \[\int{E\cdot ds}=\phi \] where \[E\] is electric field intensity, \[s\] the surface area, \[\phi \] 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. \[\phi =E\times \frac{\pi {{d}^{2}}}{4}\] \[\Rightarrow \] \[E=\frac{4\phi }{\pi {{d}^{2}}}\]
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