Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Spring 2005

Solution to øving 2

Guidance January 20. and 21.

Exercise 1

a) C Electrons have negative charge. An excess of N electrons thus implies a net negative charge:

Q = −Ne = −5 · 10^{13} · 1.6 · 10^{−19} C = −8 · 10^{−6} C = −8 μC

Here, μ denotes micro, i.e., 1 μC = 10^{−6} C.

b) A Here, it is sufficient to consider the direction of the partial forces that act on the third charge:

With Pytagoras, we have a distance of \sqrt{8}a between the two positive charges. Since the Coulomb force is proportional to 1/r^{2}, the force between the two positive charges becomes half the force between the negative and the positive charge. The vector sum becomes as indicated in the figure, i.e., a total force F with negative x component and positive y component.

c) C Total electric field in P is the vector sum of the contributions from the four point charges. The configuration in figure 3 yields the largest field strength. (No field contribution has a component upwards in this case.)

Exercise 2

a) With a line charge (i.e., charge pr unit length) λ, the charge dq and Q on a small length dx and on the whole rod becomes, respectively,

dq = λ dx Q = λL

b) Electric field from element dx in position x:

dE = \frac{λ dx}{4πε_{0}r^{2}} r̂ = A\frac{dx}{r^{2}} r̂

where we have introduced A = λ/4πε_{0}. From the figure we see that this vector has components

dE_{x} = −dE sin θ = −\frac{A dx}{r^{2}} sin θ dE_{y} = dE cos θ = \frac{A dx}{r^{2}} cos θ

Here, we have chosen x = 0 when θ = 0, and the sign is in agreement with the given information, i.e., θ > 0 when x > 0. We use the given hint and express dx and 1/r^{2} by the angle θ:

x = R tan θ ⇒ dx = \frac{Rdθ}{cos^{2} θ}
r = \frac{R}{cos θ} ⇒ \frac{1}{r^{2}} = \frac{cos^{2} θ}{R^{2}}
⇒ \frac{dx}{r^{2}} = \frac{dθ}{R}

The components E_{x} and E_{y} of the field E in the point P from the whole rod is found by integrating dE_{x} and dE_{y}:

E_{x} = ∫ dE_{x} = −\frac{A}{R} ∫_{θ_{2}}^{θ_{1}} sin θdθ = \frac{A}{R} |_{θ_{2}}^{θ_{1}} cos θ = \frac{λ}{4πε_{0}R} (cos θ_{1} − cos θ_{2})
E_{y} = ∫ dE_{y} = \frac{A}{R} ∫_{θ_{2}}^{θ_{1}} cos θdθ = \frac{A}{R} |_{θ_{2}}^{θ_{1}} sin θ = \frac{λ}{4πε_{0}R} (sin θ_{1} − sin θ_{2})

Comment: We might have been ”unfortunate” and started with the relation x = r sin θ, which yields dx = r cos θ dθ + sin θ dr, since both θ and r vary with x. But things work out nicely anyway: We have cos θ = R/r, i.e., r = R/ cos θ, and hence

dr = −R\frac{1}{cos^{2} θ}(− sin θ) dθ

so that

\frac{dx}{r^{2}} = \frac{r cos θ dθ + sin θ dr}{r^{2}}
= \frac{cos θ dθ · cos θ}{R} + \frac{sin θ · R sin θ dθ}{R^{2}}
= \frac{dθ}{R} (cos^{2} θ + sin^{2} θ)
= \frac{dθ}{R}

c) With P equally far from the two ends of the rod, we have θ_{1} = −θ_{2}, and therefore cos θ_{1} − cos θ_{2} = 0 and sin θ_{1} − sin θ_{2} = 2 sin θ_{1} = L/\sqrt{R^{2} + L^{2}/4} Thus:

E_{x} = 0

and

E = E_{y} = \frac{λL}{4πε_{0}R\sqrt{R^{2} + L^{2}/4}}

Far away from the rod, i.e., R ≫ L: We may then replace the square root by R because we may neglect L^{2}/4 in comparison with R^{2}. Then we obtain:

E ≃ \frac{λL}{4πε_{0}R^{2}} = \frac{Q}{4πε_{0}R^{2}}

This is the same as the electric field from a point charge Q in a distance R. Not unexpected: If we are far away, the rod essentially looks like a point charge with total charge Q = λL.

d) An infinitely long rod is achieved by letting θ_{2} → −π/2 and θ_{1} → π/2. Again we have E_{x} = 0, and hence

E = E_{y} = \frac{λ}{2πε_{0}R}

In other words: The field from an infinitely long line charge falls off inversely proportional with the distance R.

Exercise 3

a) The area of a thin ring with radius R and width dR is dA = 2πR dR, so the charge on such a ring is

dq = σ dA = 2πσR dR

The area of the disk is A = πR_{0}^{2}, so the total charge on the disk is

Q = σA = πσR_{0}^{2}

If one does not remember the area of a circular disk, one may of course determine the total charge Q by integrating dq:

Q = ∫ dq = ∫_{0}^{R_{0}} 2πσR dR = 2πσ |_{0}^{R_{0}} \frac{R^{2}}{2} = πσR_{0}^{2}

And if one also does not remember what the circumference of a ring is, the charge on the thin ring may be found by starting with a small angle dϕ and the area enclosed between R and R + dR. This area is Rdϕ · dR, and if we integrate this expression over ϕ from 0 to 2π, we get 2πR dR, which must then be the area of the thin ring with radius R and width dR.

b) We divide the disk into infinitesimally thin rings with width dR (see figure below). All points on the ring lie in the same distance r from the point on the z axis. Diagonally located points (or: areas dA) contribute to the field in such a way that the x and y components of the total field becomes zero (cf example in the lectures). The z component becomes:

dE_{z} = \frac{dQ}{4πε_{0}r^{2}} cos θ

Since r is constant around the whole ring, one may let dQ be the charge on the thin ring:

dQ = σR dR ∫_{0}^{2π} dϕ = 2πσR dR

Hence the field from the disk becomes

E_{z} = ∫ dE_{z} = \frac{1}{4πε_{0}} ∫_{0}^{R_{0}} \frac{2πσRa dR}{(a^{2} + R^{2})^{3/2}} = \frac{σa}{2ε_{0}} |_{0}^{R_{0}} \frac{(−1)}{\sqrt{a^{2} + R^{2}}} = \frac{σ}{2ε_{0}} (1 − \frac{a}{\sqrt{a^{2} + R_{0}^{2}}} )

Here, we have used

cos θ = \frac{a}{r} = \frac{a}{\sqrt{a^{2} + R^{2}}}

An alternative would have been to use the angle θ as integration variable:

tan θ = \frac{R}{a} ⇒ d(tan θ) = \frac{dθ}{cos^{2} θ} = \frac{dR}{a}
r = \frac{a}{cos θ}
∫_{0}^{R_{0}} \frac{R dR}{r^{2}} cos θ = ∫_{0}^{θ_{0}} (\frac{cos θ}{a} )^{2} a tan θ\frac{a dθ}{cos^{2} θ} cos θ = ∫_{0}^{θ_{0}} sin θ dθ
= 1 − cos θ_{0} = 1 − \frac{a}{r_{0}} = 1 − \frac{a}{\sqrt{a^{2} + R_{0}^{2}}}

where r_{0} and θ_{0} are defined in the figure above.

c) When a ≫ R_{0}, one might at first (as in 2c above) consider replacing \sqrt{a^{2} + R_{0}^{2}} with a. However, that only gives us the ”trivial” solution E_{z} = 0, whereas we are interested in the dominating non-zero contribution to E_{z}. This means that we must expand \sqrt{a^{2} + R_{0}^{2}} and include sufficiently many terms so that we end up with something which is no longer zero:

E_{z} = \frac{σ}{2ε_{0}} (1 − \frac{a}{a\sqrt{1 + \frac{R_{0}^{2}}{a^{2}}}} )

= \frac{σ}{2ε_{0}} (1 − (1 − \frac{R_{0}^{2}}{2a^{2}} + ⋯))
≃ \frac{σR_{0}^{2}}{4ε_{0}a^{2}}
= \frac{Q}{4πε_{0}a^{2}}

Here, we have used the approximation that was given in the text, (1 + α)^{−1/2} ≃ 1 − α/2, with α = R_{0}^{2}/a^{2} ≪ 1. This is the field in a distance a from a point charge Q = σA, where A = πR_{0}^{2} is the area of the disk. As expected: If we are sufficiently far away, we see no difference between a charged disk and a point charge.

In the opposite limit, a ≪ R_{0}, we may neglect the term a/\sqrt{a^{2} + R_{0}^{2}} in comparison with 1. This gives us

E_{z} = \frac{σ}{2ε_{0}}

In other words, a uniform electric field which depends neither on the distance a nor the extent R_{0} of the disk. Hence, this is the field outside an infinitely large plane with surface charge density σ. It might not be obvious to you that the field then becomes independent of the distance to the plane, but that’s the case! Of course, in practice we never have infinite planes of charge at our disposal, but this is nevertheless an important result: With a large charged plane, we generate an approximately uniform electric field near the plane, as long as we stay away from the edges of the plane. We will use this result many times later.