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