Electric Flux and Gauss’s Law Problems and Solutions

 Problem #1

An electric field with a magnitude of 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long assuming that (a) the plane is parallel to theyz plane; (b) the plane is parallel to the xy plane; (c) the plane contains the y axis, and its normal makes an angle of 40.0° with the x axis.

Answer;
Known:
electric field with a magnitude of E = 3.50 kN/C

(a) then electric flux through a rectangular plane 0.350 m wide and 0.700 m long assuming that the plane is parallel to theyz plane is

 φ_E = EA cos θ  = (3.50 x 10³)(0.350 x 0.700) cos 0⁰ = 858 N.m²/C

(b) if θ = 90⁰, then φ_E = EA cos θ  = (3.50 x 10³)(0.350 x 0.700) cos 90⁰ = 0

(c) if θ = 40⁰, then φ_E = EA cos θ  = (3.50 x 10³)(0.350 x 0.700) cos 40⁰ = 657 N.m²/C

Problem #2
A 40.0-cm-diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.20 x 10⁵ N.m²/C. What is the magnitude of the electric field?

Answer
Known:
The flux in this position is measured to be 5.20 x 10⁵ N.m²/C
Area, A = πr² = π(0.200)² = 0.126 m², then

φ_E = EA cos θ 

5.20 x 10⁵ = E(0.126) cos 0⁰

E =  4.14 x 10⁶ N/C

Problem #3
What is the electric flux through a sphere that has a radius of 1.0 m and carries a charge of +1.0 μ C at its center?

Answer
The magnitude of the electric field is giving by

E = kq/r² = (8.99 x 10⁹ N.m²/C²)(1.00 x 10^-6 C)/(1.00 m)² = 8.99 x 10³ N/C

The field points radially outward and is therefore everywhere perpendicular to the surface of the sphere. The flux through the sphere (whose surface area A = 4 πr² = 12.6 m²) is thus

φ_E = EA = (8.99 x 10³ N/C)( 12.6 m²) = 1.13 x 10⁵ N.m²/C

Problem #4
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 75.0 N/C that is directed at 20⁰ from the plane of the sheet. Find the magnitude of the electric flux through the sheet.

Answer;
Known:
electric field with a magnitude of E = 75.0 N/C
Area, A = 0.400 x 0.600 = 0.240  m², then the magnitude of the electric flux through the sheet is given by

φ_E = EA cos θ

φ_E = (75.0 N/C)( 0.240  m²)  cos 20⁰ = 16.91 N.m²/C

Problem #5
A carton with an area of 1.6 m² is rotated with an east-west axis in an area that has a homogeneous electric field with horizontal E = 6 x 10⁵ N/C north. Determine the electric flux that penetrates the plane of the carton when the carton is in position: (a) horizontal, (b) vertical and (c) sloping at an angle π/6 rad towards the north.

Answer
Known:
Electric field strength E = 6 x 10⁵ N / C, carton area A = 1.6 m²

(a) For a carton in a horizontal position (figure i), the angle between E and the normal direction of the plane n, which is θ = 90⁰, cos 90⁰ = 0, so that the electric flux Φ is:

Φ = EA cos θ = 0

(b) for carton that are in a vertical position (figure ii), the angle between E and n is 0⁰, cos θ = 1, so that the electric flux Φ is

Φ = EA cos θ = (6 x 10⁵ N/C)(1.6 m²)

     = 9.6 x 10⁵ Wb

(c) for cartons in an inclined position to form a 30⁰ angle to the carton (figure iii), the angle between E and n is θ = 60⁰, cos θ = ½, so that the electric flux Φ is

Φ = EA cos θ = (6 x 10⁵ N/C) (1.6 m²) (½)

     = 4.8 x 10⁵ Wb

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