Worked examples · Sheet 2

Fluxes of the field of point charges

Theory used: Gauss's law Flux of the field of a point charge through a sphere Role of the outside charges Solid angle Outward flux through a closed surface

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A closed surface $S$ encloses the point charges $q_{1} = + 3.0 nC$ and $q_{2} = - 5.0 nC$ , while the charge $q_{3} = + 4.0 nC$ is outside. Compute the outward electric flux through $S$ . Then compute the flux after the charge $q_{3}$ has been moved inside $S$ .

Φ_{E} = \frac{q _{int}}{ε _{0}} = \frac{q _{1} + q _{2}}{ε _{0}}, Φ_{E}^{'} = \frac{q _{1} + q _{2} + q _{3}}{ε _{0}}

Φ_{E} Φ_{E}^{'} = \frac{( 3.0 - 5.0 ) \cdot 1 0 ^{- 9}}{8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx - 226 N \cdot m^{2} /C = \frac{( 3.0 - 5.0 + 4.0 ) \cdot 1 0 ^{- 9}}{8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx + 226 N \cdot m^{2} /C

By Gauss’s law the outward flux through a closed surface depends only on the charge it encloses, whatever the shape of the surface. In the first situation the enclosed charge is $q_{1} + q_{2} = - 2.0 nC$ , so the outward flux is negative, that is, the field lines entering $S$ outnumber those leaving it.

The outside charge $q_{3}$ contributes to the field at every point of $S$ , but its contribution to the total flux is zero. When $q_{3}$ enters $S$ , the enclosed charge becomes $+ 2.0 nC$ and the flux changes sign, keeping the same magnitude.

Exercise 2

A point charge $q = 2.0 μ C$ is at the center of a cube. Compute the outward electric flux through one face of the cube and the outward flux through the whole cube.

Φ_{cube} = \frac{q}{ε _{0}}, Φ_{face} = \frac{1}{6} Φ_{cube} = \frac{q}{6 ε _{0}}

Φ_{cube} Φ_{face} = \frac{2.0 \cdot 1 0 ^{- 6}}{8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx 2.3 \cdot 1 0^{5} N \cdot m^{2} /C = \frac{2.0 \cdot 1 0 ^{- 6}}{6 \cdot 8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx 3.8 \cdot 1 0^{4} N \cdot m^{2} /C

The charge is inside the cube, so by Gauss’s law the outward flux through the cube is $\frac{q}{ε _{0}}$ . The six faces are equal and arranged in the same way with respect to the charge, because a rotation that takes one face to another leaves the center of the cube fixed; therefore the six fluxes are equal, and each is one sixth of the total flux. The result does not depend on the edge of the cube.

Computing the flux through one face directly would require a surface integral, because on the face the field changes magnitude and direction from point to point. The symmetry makes it possible to avoid it.

Exercise 3

A point charge $q = 2.4 nC$ is at a vertex of a cube. Compute the electric flux through each face of the cube and the total outward flux.

Φ_{large} = \frac{q}{ε _{0}} = 24 Φ_{f} ⟹ Φ_{f} = \frac{q}{24 ε _{0}}

Φ_{adjacent} = 0, Φ_{cube} = 3 Φ_{f} = \frac{q}{8 ε _{0}}

Φ_{f} Φ_{cube} = \frac{2.4 \cdot 1 0 ^{- 9}}{24 \cdot 8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx 11 N \cdot m^{2} /C = \frac{2.4 \cdot 1 0 ^{- 9}}{8 \cdot 8.85 \cdot 1 0 ^{- 12}} N \cdot m^{2} /C \approx 34 N \cdot m^{2} /C

The charge lies on the surface of the cube, so Gauss’s law does not apply directly to the cube. One then considers the eight equal cubes that share that vertex, and that together form a cube of twice the edge with the charge at its center. By Gauss’s law the outward flux through the large cube is $\frac{q}{ε _{0}}$ .

The surface of the large cube is formed by $24$ faces of the small cubes, three for each small cube, namely the three faces that do not contain the vertex. These faces are all arranged in the same way with respect to the charge, so they are crossed by the same flux $\frac{q}{24 ε _{0}}$ . The three faces of the original cube that contain the vertex lie on planes through the charge, so the field, which is radial, is parallel to them and their flux is zero. The outward flux through the cube is therefore $\frac{3 q}{24 ε _{0}} = \frac{q}{8 ε _{0}}$ , one eighth of the total flux, because the cube occupies one eighth of the directions leaving the charge.

Exercise 4

A point charge $q = 5.0 nC$ is on the axis of a disk of radius $a = 0.30 m$ , at distance $d = 0.40 m$ from its center. Compute the flux of the field of the charge through the disk, oriented with the normal pointing away from the charge.

The disk of radius

a

and the spherical cap centered at

q

with the same boundary are crossed by the same flux

R_{0} = d^{2} + a^{2}, cos α = \frac{d}{R _{0}}, A_{cap} = \int_{0}^{α} 2 π R_{0} sin φ \cdot R_{0} d φ = 2 π R_{0}^{2} (1 - cos α)

Φ_{disk} = Φ_{cap} = \frac{q}{4 π ε _{0} R _{0}^{2}} \cdot 2 π R_{0}^{2} (1 - cos α) = \frac{q}{2 ε _{0}} (1 - \frac{d}{d ^{2} + a ^{2}})

R_{0} Φ_{disk} = 0.4 0^{2} + 0.3 0^{2} m = 0.50 m, cos α = 0.80 = \frac{5.0 \cdot 1 0 ^{- 9}}{2 \cdot 8.85 \cdot 1 0 ^{- 12}} \cdot (1 - 0.80) N \cdot m^{2} /C \approx 56 N \cdot m^{2} /C

On the disk the field changes magnitude and direction, so it is convenient to replace the disk with a surface on which the field is simple. The spherical cap centered at the charge, of radius $R_{0}$ , has the same boundary as the disk; the disk and the cap bound a region that contains no charge. By Gauss’s law the outward flux through that region is zero, therefore the flux entering through the disk equals the flux leaving through the cap, that is, the flux through the disk oriented as required equals the flux through the cap oriented outward.

On the cap the field is radial, hence perpendicular to the surface, and has constant magnitude $\frac{q}{4 π ε _{0} R _{0}^{2}}$ ; the flux is the product of this magnitude and the area of the cap. The area is obtained by summing the strips between the angles $φ$ and $φ + d φ$ from the axis, which have radius $R_{0} sin φ$ and width $R_{0} d φ$ . The radius $R_{0}$ cancels, and the flux depends only on the angle $α$ , that is, on the solid angle $2 π (1 - cos α)$ under which the disk is seen from the charge.

With $cos α = 0.80$ one tenth of the total flux $\frac{q}{ε _{0}} \approx 565 N \cdot m^{2} /C$ passes through the disk. As a check, when the radius of the disk grows without bound $cos α$ tends to zero and the flux tends to $\frac{q}{2 ε _{0}}$ , half of the total, because an infinite plane divides the directions leaving the charge into two equal parts.