Electric flux – problems and solutions

1. A uniform electric field E = 8000 N/C passing through a flat square area A = 10 m². Determine the electric flux.

Known :

The magnitude of the electric field (E) = 8000 N/C

Area (A) = 10 m²

θ = 0^o (the angle between the electric field direction and a line drawn a perpendicular to the area)

Wanted: Electric flux (Φ)

Solution :

The formula of electric flux :

Φ = E A cos q

Φ = electric flux (Nm²/C), E = electric field (N/C), A = area (m²), q = angle between electric field line with the normal line.

Electric flux :

Φ = E A cos q = (8000)(10)(cos 0) = (8000)(10)(1) = 80,000 = 8 x 10⁴ Nm²/C

2. A uniform electric field E = 5000 N/C passing through a flat square area A = 2 m². Determine the electric flux.

Known :

Electric field (E) = 5000 N/C

Area (A) = 2 m²

θ = 60^o (the angle between the electric field direction and a line drawn perpendicular to the area)

Wanted : Electric flux (Φ)

Solution :

Electric flux :

Φ = E A cos q = (5000)(2)(cos 60) = (5000)(2)(0.5) = 5000 = 5 x 10³ Nm²/C

3. A solid ball with 0.5 meters radius has 10 μC electric charge in its center. Determine the electrical flux pass through the solid ball.

Known :

Radius of ball (r) = 0.5 m

Electric charge (Q) = 10 μC = 10 x 10^-6 C

Wanted : Electric flux (Φ)

Solution :

Electric field :

E = k q/r²

E = (9 x 10⁹ Nm²/C²)(10 x 10^-6 C) / 0.5²

E = (90 x 10³) / 0,25

E = 360 x 10³

E = 3.60 x 10⁵ N/C

Surface area :

A = 4 π r² = 4 (3.14)(0.5)² = (12.56)(0.25) = 3.14 m²

Electric flux :

The electric field lines perpendicular to area, so that the angle between the electric field direction and a line drawn perpendicular to the area, is 0^o.

Φ = E A cos q

Φ = (3.60 x 10⁵)(3.14)(cos 0)

Φ = (11.304 x 10⁵)(1)

Φ = 11.304 x 10⁵

Φ = 1.13 x 10⁶ Nm²/C