Find the flux of the electric field through surface ABCD of the inclined plane as shown in figure.
A
Eabcosθ
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B
Eabsinθ
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C
Eabtanθ
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D
Eabcotθ
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Solution
The correct option is BEabsinθ Given: AB=a;BC=b
So, area of surface ABCD,A=ab
From above figure,
angle (α) between electric field lines and area vector is (90∘−θ)