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Question
The electric field in a region is given by `vec"E"` = 5 `hatk`N/C. Calculate the electric flux Through a square of side 10.0 cm in the following cases
- The square is along the XY plane
- The square is along XZ plane
- The normal to the square makes an angle of 45° with the Z axis.
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Solution
Given: `vec"E"` = 5 `hatk` N/C, |E| = 5 N/C, l = 10 cm = 10 × 10−2 m = 10−1 m, A = l2 = 10−2m2
To find: Electric flux in three cases (Φ1), (Φ2), (Φ3)
(i) Given the plane of Area is along XY plane. hence the direction of its normal `(vec"dS")` vector will be along Z axis.
The Angle between `vec"E"` and `"d"vec"S"` is zero i.e. θ = 0°.
∴ Φ = EA cos θ
∴ Φ1 = 5 × 10−2 cos 0
∴ Φ1 = 5 × 10−2 V m
(ii) Plane of Area is along XZ plane. Hence the direction of its normal `(vec"dS")` vector will be along Y axis.
∴ θ = 90°
∴ Φ = EA cos θ
∴ Φ2 = 5 × 10−2 cos 90°
∴ Φ2 = 0 V m
(iii) When normal to the square makes an angle of 45° with the Z axis.
∴ θ = 45°
∴ Φ = EA cos θ
∴ Φ3 = 5 × 10−2 cos 45°
∴ Φ3 = 3.5 × 10−2 V m
Electric flux in the above-mentioned cases are 5 × 10−2 V m, 0 V m, and 3.5 × 10−2 V m.
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