Answer:
0 Nm²/C
Explanation:
We calculate the total flux Ф = ∫E.dA for -15 < x < 0 and 0 < x < 15
which is the flux through the curved sides + the flux through its ends.
So, Ф₁ = ∫₋₁₅⁰E.dA + ∫₀¹⁵E.dA this is the flux through the curved sides of the cylinder
Ф₁ = ∫₋₁₅⁰EdAcos0 + ∫₀¹⁵EdAcos0 (since the area is parallel to the electric field)
Ф₁ = ∫₋₁₅⁰EdA + ∫₀¹⁵EdA
Ф₁ = E∫₋₁₅⁰dA + E∫₀¹⁵dA
= E∫₋₁₅⁰2πrdx + E∫₀¹⁵2πrdx
= E2πr∫₋₁₅⁰dx + E2πr∫₀¹⁵dx
= E2πr[x]₋₁₅⁰ + E2πr[x]¹⁵₀
= E2πr[0 -(-15)] + E2πr[15 - 0]
= (-400 N/C)2πr[15] + (+400 N/C)2πr[15]
= -800 N/Cπ(0.30 m )[0.15 m] + 800 N/Cπ(0.30 m)[0.15]
= (-36επ + 36επ) Nm²/C = 0
the flux through the ends is
Ф₂ = ∫₋₁₅⁰E.dA + ∫₀¹⁵E.dA
= ∫₋₁₅⁰EdAcos0 + ∫₀¹⁵EdAcos0 (since the area is parallel to the electric field)
= ∫₋₁₅⁰EdA + ∫₀¹⁵EdA
= E∫₋₁₅⁰dA + E∫₀¹⁵dA
= (-400 N/C)πr² + (400 N/C)πr²
= 0
So, the total flux equals Ф = Ф₁ + Ф₂ = 0 + 0 = 0 Nm²/C
