Electrostatics Question 731

Question: A loop of diameter d is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $ \phi . $ What is the electric field strength?

Options:

A) $ \frac{4\phi }{\pi d^{2}} $

B) $ \frac{2\phi }{\pi d^{2}} $

C) $ \frac{\phi }{\pi d^{2}} $

D) $ \frac{\pi \phi d^{2}}{4} $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $ \phi =EA\cos 0{}^\circ =E\times \frac{\pi d^{2}}{4},\therefore E=\frac{4\phi }{\pi d^{2}} $



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