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प्रश्न
Consider a uniform electric field E = 3 × 103 `bbhat i` N/C.
- What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
- What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
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उत्तर
- Area of square, A= a2 = 102 = 100 cm2
= 100 × 10-4 m2
= 10-2 m2
Since the plane lies on the y-z-axis, area vector `vecA` points in the same direction.
i.e., `vecA = (10^-2 hati) "m"^2`
∴ Electric Flux through the square is
Φ = `vecE · vecA`
= `(3 xx 10^3 hati) · (10^-2 hati)`
= 3 × 101
or Φ = 30 V-m - When normal to plane i.e., A makes an angle of 60° with E, then
Φ' = EA cos 60°
= 3 × 103 × 10-2 × 1/2
= 1.5 × 101
or Φ' = 15 V-m
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| (i) | (ii) | (iii) | (iv) |
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