# Electric flux – problems and solutions

1. A uniform electric field E = 8000 N/C passing through a flat square area A = 10 m^{2}. Determine the electric flux.

__Known :__

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

Area (A) = 10 m^{2}

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 (**Nm*^{2}*/C)**, E = electric field (N/C), A = area (m*^{2}*), **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 10^{4} Nm^{2}/C

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

__Known :__

Electric field (E) = 5000 N/C

Area (A) = 2 m^{2}

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 10^{3} Nm^{2}/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/r^{2 }

E = (9 x 10^{9} Nm^{2}/C^{2})(10 x 10^{-6} C) / 0.5^{2}

E = (90 x 10^{3}) / 0,25

E = 360 x 10^{3}

E = 3.60 x 10^{5} N/C

Surface area :

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

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 10^{5})(3.14)(cos 0)

Φ = (11.304 x 10^{5})(1)

Φ = 11.304 x 10^{5}

Φ = 1.13 x 10^{6} Nm^{2}/C