An electric field is = (400 N/C) for x > 0 and = (–400 N/C) for x < 0. A cylinder of length 30 cm and radius 10 cm has its center at the origin and its axis along the x axis such that one end is at x = +15 cm and the other is at x = –15 cm. What is the net outward flux through the entire cylindrical surface?

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

6/12=1/2

step-by-step explanation: