Electrostatics Question 729
If the electric flux entering and leaving an enclosed surface respectively is $ {\phi _{1}} $ and $ {\phi _{2}} $ , the electric charge inside the surface will be
Options:
A) $ ( {\phi _{2}}+{\phi _{2}} )\times {\varepsilon _{0}} $
B) $( {\phi _{2}}-{\phi _{1}} )\times {\varepsilon _{0}} $
C) $ ( {\phi _{1}}+{\phi _{2}} )\times {\varepsilon _{0}} $
D) $ ( {\phi _{1}}-{\phi _{2}} )\times {\varepsilon _{0}} $
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Answer:
Correct Answer: D
Solution:
[d] $ \phi =EA\cos 0{}^\circ =E\times \frac{\pi d^{2}}{4},\therefore E=\frac{4\phi }{\pi d^{2}} $
BETA
