Answer:
Electric flux;
Φ = 30.095 × 10⁴ N.m²/C
Explanation:
We are given;
Charge on plate; q = 17 µC = 17 × 10^(-6) C
Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²
Angle between the normal of the area and electric field; θ = 4°
Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m
Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²
The charge density on the plate is given by the formula;
σ = q/A_p
Thus;
σ = (17 × 10^(-6))/(180 × 10^(-4))
σ = 0.944 × 10^(-3) C/m²
Also, the electric field is given by the formula;
E = σ/ε_o
E = (0.944 × 10^(-3))/(8.85 × 10^(-12))
E = 1.067 × 10^(8) N/C
Now, the formula for electric flux for uniform electric field is given as;
Φ = EAcos θ
Where A = πr² = π × 0.03² = 9π × 10^(-4) m²
Thus;
Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4
Φ = 30.095 × 10⁴ N.m²/C
Answer:
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