Worked examples · Sheet 1

Fluxes through flat and closed surfaces

Theory used: Electric flux Outward flux through a closed surface Flux and field lines Gauss's law

Exercise 1

A rectangular surface with sides $a = 0.20 m$ and $b = 0.30 m$ is in a uniform electric field of magnitude $E = 4.0 \cdot 1 0^{3} N/C$ , and its normal $\overset{n}{^}$ makes the angle $θ = 6 0^{\circ}$ with the field. Compute the flux of the field through the surface, the maximum flux obtained by rotating it and the orientation for which the flux is zero.

Φ_{E} = E \cdot \overset{n}{^} A = E ab cos θ, Φ_{m a x} = E ab (θ = 0)

Φ_{E} Φ_{m a x} = 4.0 \cdot 1 0^{3} \cdot 0.20 \cdot 0.30 \cdot cos 6 0^{\circ} N \cdot m^{2} /C = 4.0 \cdot 1 0^{3} \cdot 0.060 \cdot 0.50 N \cdot m^{2} /C = 1.2 \cdot 1 0^{2} N \cdot m^{2} /C = 4.0 \cdot 1 0^{3} \cdot 0.060 N \cdot m^{2} /C = 2.4 \cdot 1 0^{2} N \cdot m^{2} /C

The surface is flat and the field is uniform, so $E \cdot \overset{n}{^}$ is the same at every point and by the definition of electric flux the integral reduces to the product $E A cos θ$ , with $A = ab = 0.060 m^{2}$ . With $θ = 6 0^{\circ}$ the cosine equals $0.50$ , and the flux is $1.2 \cdot 1 0^{2} N \cdot m^{2} /C$ .

Rotating the surface changes only the angle $θ$ . The cosine is maximum for $θ = 0$ , that is, when the normal is parallel to the field and the surface is perpendicular to the field lines, and the maximum flux is twice the one computed. The flux is zero for $θ = 9 0^{\circ}$ , that is, when the surface is parallel to the field and no field line crosses it.

Exercise 2

A wedge shaped like a right prism has as its bases two right triangles with horizontal leg $a = 0.40 m$ and vertical leg $h = 0.30 m$ , and its horizontal edges have length $w = 0.50 m$ . The wedge rests on a horizontal plane on its rectangular face with sides $a$ and $w$ , and it is immersed in a uniform horizontal field of magnitude $E = 2.5 \cdot 1 0^{3} N/C$ , perpendicular to the horizontal edges and directed from the vertical face toward the inclined face. Compute the outward flux through each face and the total outward flux.

Cross-section of the wedge with a plane perpendicular to the horizontal edges; the field enters through the vertical face and leaves through the inclined face

E = E \overset{}{^}, \overset{n}{^}_{1} = - \overset{}{^}, \overset{n}{^}_{2} = \frac{h  ^ + a  ^}{ℓ} with ℓ = a^{2} + h^{2}, \overset{n}{^}_{3} = - \overset{}{^}, \overset{n}{^}_{4, 5} = \pm \hat{k}

Φ_{1} = - E h w, Φ_{2} = E \frac{h}{ℓ} ℓ w = E h w, Φ_{3} = Φ_{4} = Φ_{5} = 0, Φ_{tot} = 0

ℓ Φ_{1} Φ_{2} = 0.4 0^{2} + 0.3 0^{2} m = 0.50 m = - 2.5 \cdot 1 0^{3} \cdot 0.30 \cdot 0.50 N \cdot m^{2} /C \approx - 3.8 \cdot 1 0^{2} N \cdot m^{2} /C = + 2.5 \cdot 1 0^{3} \cdot 0.30 \cdot 0.50 N \cdot m^{2} /C \approx + 3.8 \cdot 1 0^{2} N \cdot m^{2} /C

Choose the $x$ axis in the direction of the field, the $y$ axis vertical and pointing up, and the $z$ axis along the horizontal edges. Each face is flat and the field is uniform, so the outward flux through each face is the dot product $E \cdot \overset{n}{^}$ times the area, with $\overset{n}{^}$ the unit normal pointing outward from the wedge.

The vertical face has area $h w$ and outward normal $- \overset{}{^}$ , opposite to the field, therefore its flux is negative, and the field enters the wedge through that face. The inclined face has area $ℓ w$ , and its outward normal is perpendicular to the hypotenuse of the base triangle; since the hypotenuse has the direction of the vector $(a, - h)$ , the outward normal has the direction of $(h, a)$ , and the component along the field is $\frac{h}{ℓ} = 0.60$ . The flux is therefore $E \cdot 0.60 \cdot ℓ w = E h w$ , equal to the flux entering through the vertical face. The horizontal base and the two triangles have their normal perpendicular to the field, so their flux is zero.

The total outward flux is zero, as it must be for a closed surface that encloses no charge. The computation also shows that the flux through the inclined face equals the flux through its projection $h w$ onto the plane perpendicular to the field, because the same field lines pass through the two surfaces.

Exercise 3

In a region of space the electric field is $E = b x \overset{}{^}$ , with $b = 4.0 \cdot 1 0^{3} N/ (C \cdot m)$ . A cube of edge $a = 0.50 m$ has one vertex at the origin and its edges along the positive $x$ , $y$ , $z$ semi-axes. Compute the outward flux through the cube and the charge contained in the cube.

Φ_{x = a} = E_{x} (a) a^{2} = b a^{3}, Φ_{x = 0} = - E_{x} (0) a^{2} = 0, Φ_{others} = 0

Φ_{E} = b a^{3}, Φ_{E} = \frac{q _{int}}{ε _{0}} ⟹ q_{int} = ε_{0} b a^{3}

Φ_{E} q_{int} = 4.0 \cdot 1 0^{3} \cdot 0.5 0^{3} N \cdot m^{2} /C = 5.0 \cdot 1 0^{2} N \cdot m^{2} /C = 8.85 \cdot 1 0^{- 12} \cdot 5.0 \cdot 1 0^{2} C \approx 4.4 \cdot 1 0^{- 9} C

The field has only the $x$ component, so it is parallel to the four faces perpendicular to the $y$ and $z$ axes, and the flux through these faces is zero. On the two faces perpendicular to the $x$ axis the field does not depend on $y$ and $z$ , therefore on each of them it is constant and the flux is the product of the normal component and the area $a^{2}$ . On the face $x = 0$ the field is zero; on the face $x = a$ it equals $b a$ and points in the direction of the outward normal $\overset{}{^}$ , so the flux is $b a \cdot a^{2}$ .

The field is not uniform, and the total outward flux is nonzero. By Gauss’s law the charge contained in the cube is $ε_{0}$ times the flux, about $4.4 nC$ . As a dimensional check, $b a^{3}$ is measured in $N/ (C \cdot m) \cdot m^{3} = N \cdot m^{2} /C$ , the unit of flux.