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}} $
