Institutt for fysikk, NTNU TFY4155/FY1303: Elektrisitet og magnetisme Vår 2004

Øving 5

Guidance: Monday February 9 To be delivered by: Thursday February 12

Exercise 1

In exercise 2 in øving 3, we investigated an electric dipole, consisting of two point charges ±q located on the z axis in z = ±a/2. We showed that the potential V far away (r ≫ a) from the dipole is approximately equal to

V (r, θ) = \frac{p cos θ}{4πε_{0}r^{2}}

Here, r is the distance form the origin, i.e., the centre of the dipole, θ is the angle between the z axis and r, and p = |p| = qa is the electric dipole moment of the dipole.

a) Starting from the expression above for V (r, θ), determine the electric field E(r, θ) = E_{r}r̂+E_{θ}θ̂ far away from the dipole. The gradient operator in spherical coordinates is

∇ = r̂\frac{∂}{∂r} + θ̂\frac{1}{r} \frac{∂}{∂θ} + ϕ̂\frac{1}{r sin θ} \frac{∂}{∂ϕ}

I will not provide the answer, but you can to some extent control your answer by checking that the result is reasonable for θ = 0 and for θ = π/2. What about r = 0?

b) Because of rotational symmetry around the z axis, we may e.g. assume that we are in the xz plane. Determine the electric field E(x, z) = E_{x}x̂ + E_{z}ẑ in cartesian coordinates for r ≫ a. Hint: Start with the expressions you found for E_{r} amd E_{θ} in a). Make a figure and find the relation between the coordinates (x, z) and (r, θ), and the components of the electric field, E_{x}, E_{z} and E_{r}, E_{θ}. [Answer: E_{x} = 3pxz/4πε_{0}(x^{2} + z^{2})^{5/2}, E_{z} = p(2z^{2} − x^{2})/4πε_{0}(x^{2} + z^{2})^{5/2}.] c) Also find E(x, z) by first rewriting V (r, θ) in cartesian coordinates, and then using the gradient operator in cartesian coordinates on V (x, z).

Exercise 2

The figure above shows a gaussian surface (i.e., a closed surface) S formed as a cube with sides a. The surface is located in a region where there exists an electric field E. In each of the cases a) – d) below, determine the total (net) electric flux ϕ that passes through the surface S. Then use Gauss’ law and find in each case also the total charge Q inside S.

a) E = Cx̂
b) E = Cxx̂
c) E = Cx^{2}x̂
d) E = C (yx̂ + xŷ)

Here, C is a (scalar) constant (with different units in the different cases, of course).

e) For c) above, determine the charge density (i.e., charge pr unit volume) ρ inside S. Hint: Use Gauss’ law with a gaussian surface enclosing a thin slice with thickness dx and top and bottom surfaces with area a^{2}, located between x and x + dx. (So the volume of the slice is a^{2} dx.)

Some answers: b): Q = Cε_{0}a^{3}  c): Q = Cε_{0}a^{4}  e): ρ = 2Cε_{0}x

Exercise 3

Use Gauss’ law and find the electric field in a distance r from an infinitely long (thin) rod with charge λ pr unit length. Hint: Take advantage of the cylindrical symmetry of the system and find a useful gaussian surface. (Compare your result with what you found in exercise 2 d) in øving 2.)

Exercise 4 (multiple choice)

a) On a closed surface, the electric field E is everywhere directed inwards. Then we may conclude that

A the surface normal n̂ is parallel with E everywhere on the closed surface
B the surface encloses zero net charge
C the surface encloses a negative net charge
D the surface encloses a positive net charge

b) The figure illustrates a closed surface enclosing two point charges q og −q. The net electric flux out through this surface is then

A zero
B −q/ε_{0}
C q/ε_{0}
D 2q/ε_{0}

c) The figure illustrates a closed surface enclosing two point charges −2q og q. The net electric flux out through this surface is then

A zero
B −q/ε_{0}
C q/ε_{0}
D 2q/ε_{0}

d) What is the radius of a (spherical) equipotential surface at 50 V with a point charge 10 nC in the centre? (Zero potential is chosen at infinity.)

A 1.3 m
B 1.8 m
C 3.2 m
D 5.0 m

e) The potential in a region of space is V (x, y, z) = 100 V. The electric field E in this region is then

A (100 V/m) x̂
B (100 V/m) ŷ
C (100 V/m) ẑ
D zero

f) A uniformly charged infinitely large surface has a charge σ pr unit area. Three gaussian surfaces (closed surfaces) a, b, and c are shown in the figure. All three surfaces enclose a circular disc with radius R when they cut through the charged surface. Range the three closed surfaces a, b, and c with respect to how much net electric flux that passes out through them.

\begin{matrix}A & a > b > c \\ B & a > b = c \\ C & a = b = c \\ D & a < b < c \end{matrix}

g) If the potential V as a function of the distance r from a charge distribution is as given in graph nr 1, which graph then shows the electric field E as a function of the distance r?

A 2
B 3
C 4
D 5

h) The potential in a region is

V (x) = 50 V + (15 V/m)x

The electric field in this region is then

A 50 V x̂
B (15 V/m) x x̂
C (15 V/m) x̂
D –(15 V/m) x̂

i) The potential in a region is

V (x, y, z) = (2 V/m)x + (3 V/m)y + (4 V/m)z

Then the x component of the electric field in this region is

A –2 V/m
B –3 V/m
C –4 V/m
D –9 V/m

j) A point charge q is located in one of the corners of a cube. What is the electric flux through the shaded side in the figure?

\begin{matrix}A & q/ε_{0} \\ B & q/4ε_{0} \\ C & q/8ε_{0} \\ D & q/24ε_{0} \end{matrix}