Fluxus electricus – problemata et solutiones

1. Campus electricus uniformis E = 8000 N/C per aream quadratam planam A = 10 m² transit ^. Fluxum electricum determina.

Notum:

Magnitudo campi electrici (E) = 8000 N/C

Area (A) = 10 ^m²

θ = 0 ^° (angulus inter directionem campi electrici et lineam perpendiculariter ad aream ductam)

Quaesitum: Fluxus electricus ( Φ)

solution:

Formula fluxus electrici:

Φ = EA cos q

Φ = fluxus electricus ( Nm² ^/ C) , E = campus electricus (N/C), A = area (m² ⁾ , q = angulus inter lineam campi electrici et lineam normalem.

Fluxus electricus:

Φ = EA cos q = (8000)(10)( cos 0) = (8000)(10)(1 ) = 80,000 = 8 x 10 ⁴ Nm ² /C

[irp]

2. Campus electricus uniformis E = 5000 N/C per aream quadratam planam A = 2 m² transit ^. Fluxum electricum determina.

Notum:

Campus electricus (E) = 5000 N/C

Area (A) = 2 ^m²

θ = 60 ^° (angulus inter directionem campi electrici et lineam perpendiculariter ad aream ductam)

Quaesitum: Fluxus electricus ( Φ)

solution:

Fluxus electricus:

Φ = EA cos q = (5000)(2)( cos 60) = (5000)(2)(0.5 ) = 5000 = 5 x 10 ³ Nm ² /C

[irp]

3. Sphaera solida cum radiis 0.5 metrorum decem μC electricitatem in centro habet. Fluxum electricum per phaeram solidam transeuntem determina.

Notum:

Radius pilae (r) = 0.5 m

Carica electrica (Q) = 10 μC = 10 × ^10⁻⁶ C

Quaesitum: Fluxus electricus ( Φ)

solution:

Campus electricus:

E = kq/ ^r²

E = (9 × 10⁶ ^{Nm² /} C² ) ⁽ 10 × ^{10⁻⁶ C} ) / 0.5²

E = (90 × ^10³ ) / 0,25

E = 360 × ^10³

E = 3.60 × 10⁵ ^N /C

Area superficiei:

A = 4 π r ² = 4 (3.14)(0.5) ² = (12.56)(0.25) = 3.14 m ²

Fluxus electricus:

Lineae campi electrici perpendiculares areæ sunt, ita ut angulus inter directionem campi electrici et lineam perpendiculariter areæ ductam sit 0 ^°.

Φ = EA cos q

Φ = ( 3.60 x 10 ⁵ )( 3.14 )( cos 0 )

Φ = (11.304 × 10⁻⁵ )( ¹ )

Φ ⁼ 11.304 × 10⁵

Φ = 1.13 × 10⁶ ^{Nm² /} C