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

Solution to øving 4

Guidance Monday February 2

Exercise 1 The potential difference ΔV between two points in space is given by

ΔV = V_{B} − V_{A} = − ∫_{A}^{B} E · dl

In this exercise, we have a uniform electric field E = E_{0} x̂, so we may write

ΔV = −E_{0} x̂ · ∫_{A}^{B} dl

where A represents the origin, (0, 0), and B the three points given in the text. We find: (i)

∫_{A}^{B} dl = ∫_{(0,0)}^{(a,0)} dl = a x̂

hence

ΔV = −E_{0} x̂ · a x̂ = −E_{0}a

(ii)

∫_{A}^{B} dl = ∫_{(0,0)}^{(0,a)} dl = a ŷ

hence

ΔV = −E_{0} x̂ · a ŷ = 0

(iii)

∫_{A}^{B} dl = ∫_{(0,0)}^{(a,a)} dl = a x̂ + a ŷ

hence

ΔV = −E_{0} x̂ · (a x̂ + a ŷ) = −E_{0}a

Exercise 2

a) With our choice for the polar angle θ, we see from the figure that

x = r sin θ
z = r cos θ
r = \sqrt{x^{2} + z^{2}}

b) We use the super position principle to determine the potential from the two point charges. With the point (x, z) in a distance r_{1} from q and r_{2} from −q, we have

V (x, z) = \frac{q}{4πε_{0}r_{1}} − \frac{q}{4πε_{0}r_{2}}
= \frac{q}{4πε_{0}} (\frac{1}{\sqrt{x^{2} + (z − a/2)^{2}}} − \frac{1}{\sqrt{x^{2} + (z + a/2)^{2}}} )

The distances r_{1} and r_{2} in terms of x and z are found directly by looking at the figure.

The potential on the x axis is

V (x, 0) = \frac{q}{4πε_{0}} (\frac{1}{\sqrt{x^{2} + a^{2}/4}} − \frac{1}{\sqrt{x^{2} + a^{2}/4}}) = 0

The potential on the z axis is

V (0, z) = \frac{q}{4πε_{0}} (\frac{1}{|z − a/2|} − \frac{1}{|z + a/2|})

Note that we must take absolute values in order to have one expression valid throughout the z axis. With z > a/2:

\frac{1}{|z − a/2|} − \frac{1}{|z + a/2|} = \frac{1}{z − a/2} − \frac{1}{z + a/2} = \frac{a}{z^{2} − a^{2}/4}

With z < −a/2:

\frac{1}{|z − a/2|} − \frac{1}{|z + a/2|} = −\frac{1}{z − a/2} + \frac{1}{z + a/2} = −\frac{a}{z^{2} − a^{2}/4}

With −a/2 < z < a/2:

\frac{1}{|z − a/2|} − \frac{1}{|z + a/2|} = −\frac{1}{z − a/2} − \frac{1}{z + a/2} = −\frac{2z}{z^{2} − a^{2}/4} = \frac{2z}{a^{2}/4 − z^{2}}

A sketch of V (0, z): c) We use the hint given in the text, in addition to the following figure, and obtain:

V (r, θ) = \frac{q}{4πε_{0}} (\frac{1}{r_{1}} − \frac{1}{r_{2}} )
= \frac{q}{4πε_{0}} · \frac{r_{2} − r_{1}}{r_{1}r_{2}}
≃ \frac{q}{4πε_{0}} · \frac{a cos θ}{r^{2}}
= \frac{p cos θ}{4πε_{0}r^{2}}
= \frac{pr cos θ}{4πε_{0}r^{3}}
= \frac{p · r}{4πε_{0}r^{3}}

Alternatively, proceeding a bit more slowly: From the figure, we see that

r_{1} ≃ r − \frac{a}{2} cos θ
r_{2} ≃ r + \frac{a}{2} cos θ

When r ≫ a we may expand both 1/r_{1} and 1/r_{2} in power series around 1/r:

\frac{1}{r_{1}} − \frac{1}{r_{2}} ≃ \frac{1}{r − \frac{a}{2} cos θ} − \frac{1}{r + \frac{a}{2} cos θ}

= \frac{1}{r} [(1 − \frac{a cos θ}{2r} )^{−1} − (1 + \frac{a cos θ}{2r} )^{−1}]

≃ \frac{1}{r} [1 + \frac{a cos θ}{2r} − 1 + \frac{a cos θ}{2r} ]
= \frac{a cos θ}{r^{2}}

Is it reasonable that the potential from an electric dipole falls off faster than the potential from a point charge (i.e., an electric “monopole”)? Yes, because the negative and the positive point charges of the dipole contribute with opposite signs to the total potential. Thus, the contributions to the potential from the two point charges partly cancel each other. (On the x axis, the two contributions cancel exactly.)

Extra, if you wonder how one should proceed in order to find the dominating correction to the result obtained above: A first thought might be to continue the series expansion that was started above, and include sufficiently many terms so that a dominating correction was obtained. If we include one more term, nothing new is obtained since that term comes with the same sign in the expansion of the two square roots, and therefore (with the minus sign in front of one of them) cancel each other. We must include two more terms:

\frac{1}{r} [(1 − \frac{a cos θ}{2r} )^{−1} − (1 + \frac{a cos θ}{2r} )^{−1}]

= \frac{1}{r} [1 + \frac{a cos θ}{2r} + (\frac{a cos θ}{2r} )^{2} + (\frac{a cos θ}{2r} )^{3} + … − (1 − \frac{a cos θ}{2r} + (\frac{a cos θ}{2r} )^{2} − (\frac{a cos θ}{2r} )^{3} + …)]

= \frac{a cos θ}{r^{2}} − \frac{a^{3} cos^{3} θ}{4r^{4}} + …
= \frac{a cos θ}{r^{2}} (1 − \frac{a^{2} cos^{2} θ}{4r^{2}} + …)

Here, we used the series expansion (1 + x)^{−1} = 1 − x + x^{2} − x^{3} + … (valid for |x| < 1). Not a bad try. However, there is a catch here: The starting point for this series expansion was an approximation itself, namely

r_{1} ≃ r − \frac{a}{2} cos θ
r_{2} ≃ r + \frac{a}{2} cos θ

And the errors we do in these approximations are of the same order of magnitude as the correction term we are looking for! The solution is obtained by going back to the exact expression for V, with r_{1} and r_{2} expressed in terms of cartesian coordinates x and z. The calculation is not difficult, but rather tedious, so the details are skipped here. If my calculation is correct, the answer is

\frac{1}{r_{1}} − \frac{1}{r_{2}} = \frac{a cos θ}{r^{2}} [1 − \frac{3a^{2}}{8r^{2}} (1 − \frac{5}{3} cos^{2} θ) + …]

Here, we have included all corrections that are an order a^{2}/r^{2} smaller than the dominating result. The next term in this series will be further reduced, by an additional factor a^{2}/r^{2}. In series expansions like this, the first term that is not included will always be smaller than the last term that we did include, since the series is a polynomial with increasing powers of a parameter which is small compared to 1. (In our particular case, we see that for directions given by cos^{2} θ ≃ 3/5, the first correction actually disappears. In that case, one may have to look at the next term in the series expansion.)

Exercise 3

a) C

F = qE = ma

Newton’s 2. law! Here, q = −e, so the acceleration of the electron becomes

a = −\frac{e}{m}E

i.e., to the left.

b) C

ΔV = − ∫ E · dl = 0

provided that

dl ⊥ E

c) D A is nonsense, V is a scalar, and it is E, not V, that is measured in units of N/C.

d) A

U = ∑_{i<j} \frac{q_{i}q_{j}}{4πε_{0}r_{ij}}
= \frac{1}{4πε_{0}l} (q_{1}q_{2} + q_{1}q_{3} + q_{2}q_{3})
= 9 · 10^{9} · \frac{1}{0.5} (50 + 75 + 150) · 10^{−6−6}
= 18 · 10^{9} · 275 · 10^{−12}
= 4.95 J

e) C Like the previous one, but with q_{2} = −10 μC ⇒

U = 18 · 10^{9} · (−50 + 75 − 150) · 10^{−12} = −2.25 J f) B

U = \frac{e^{2}}{4πε_{0}r} = e · \frac{e}{4πε_{0}r} = e · 1.6 · 10^{−19} · 9 · 10^{9} · 10^{10} = 14.4 eV

g) D Energy conservation yields

\frac{1}{2}mv^{2} = qV

I.e., acceleration of a particle with charge q and mass m through a potensial difference V results in a reduction of potential energy, qV, and a corresponding increase in kinetic energy, mv^{2}/2. Equal speed for the two particles then implies

\frac{q_{α}V_{α}}{m_{α}} = \frac{q_{Be}V_{Be}}{m_{Be}}

in other words

\frac{V_{Be}}{V_{α}} = \frac{q_{α}m_{Be}}{q_{Be}m_{α}} = \frac{2 · 9}{4 · 4} = \frac{9}{8}