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.

Electric flux through area and closed surface – problems and solutions 1Known :

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

Area (A) = 10 m2

θ = 0o (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

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

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Electric flux through area and closed surface – problems and solutions 2Known :

Electric field (E) = 5000 N/C

Area (A) = 2 m2

θ = 60o (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

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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.

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Electric flux through area and closed surface – problems and solutions 3Known :

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 0o.

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