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NCERT Class XII Chapter
Electric Charges and Fields
Questions With Solutions

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Question : 15 of 34
Marks: +1, -0
Consider a uniform electric field E→\overset{\rightarrow}{E} = 3 × 109i^NC−110^9 \hat{i} \mathrm{NC}^{-1}.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Solution:  
(a) Area of square, A = a2a^2 = 10210^2 = 100 cm2\mathrm{cm}^2 = 100 × 10−410^{-4} m2\mathrm{m}^2 = 10−2m210^{-2}\mathrm{m}^2
As the plane is along yz plane, so area vector A→\overset{\rightarrow}{A} is directed along x-axis
i.e., A→\overset{\rightarrow}{A} = (10−2i^)m2(10^{-2}\hat{i})\mathrm{m}^2
∴ Electric Flux through the square is
ϕ = E→⋅A→\overset{\rightarrow}{E} \cdot \overset{\rightarrow}{A} = (3×103i^)(3 \times 10^{\hat{3i}}) . (10−2i^)(10^{-2}\hat{i}) = 3 × 10110^1 or ϕ = 30 V-m
(b) When normal to plane i.e. A→\overset{\rightarrow}{A} makes an angle of 60° with E→\overset{\rightarrow}{E} , then ϕ' = EA cos 60° = 3 × 10310^3 × 10−210^{-2} × 1/2 = 1.5 × 10110^1 or ϕ' = 15 V-m
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