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CBSE Class 12 Physics 2014 Delhi Set 3 Paper

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Question : 6
Total: 8
Given a uniform electric field
→
E
=4×103
^
i
N∕C. Find the flux of this field through a square of 5‌cm on a side whose plane is parallel to the Y−Z plane. What would be the flux through the same square, if the plane makes an angle of 30∘ with the X-axis?
Solution:  
(a) The plane of the square is parallel to the y−z plane. Hence, angle between the unit vector normal to the plane and electric field, θ=0∘
Flux (ϕ) through the plane is given by the relation,
ϕ‌=|
→
E
|A‌cos‌θ
‌=4×103×0.25×cos‌0∘
‌=10Nm2∕C
(b) Plane makes an angle of 30∘ with the x-axis. Hence, angle between the unit vector normal to the plane and electric field, θ=60∘
Flux, ϕ‌=|
→
E
|A‌cos‌θ
‌=4×103×0.25×cos‌60∘
‌=5Nm2∕C
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