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

Solution to øving 1

Guidance January 13. and 14.

First, a comment about notation: On the blackboard, vectors are denoted with an arrow above the symbol. In written notes, like here, we typically use bold symbols. This means that F denotes the vector (e.g. a force), both its magnitude F and its direction. Unit vectors, i.e. (dimensionless) vectors of length 1, are denoted with a ˆ above the symbol. This gives us the relations F = |F |F̂ = F F̂. Further, we denote the components of a vector with subscripts, e.g., F_{x} for the x component of F.

Exercise 1

a) Correct answer is C:

A = |A| = \sqrt{A_{x}^{2} + A_{y}^{2}} = \sqrt{8.5^{2} + 1.3^{2}} = 8.6

b) Correct answer is B: The vector A has negative x component and positive y component, which means that is lies in the 2. quadrant. Let us first determine the angle θ between the negative x axis and A. We have

tan θ = \frac{|A_{y}|}{|A_{x}|} = \frac{2.3}{3.7}

which yields θ = 32 degrees. The desired angle, going counterclockwise from positive x axis towards A, becomes 180 - θ = 148 degrees.

c) Correct answer is A: Let us write C = B − A. Then

C_{x} = B_{x} − A_{x} = −10.7
C_{y} = B_{y} − A_{y} = 3.9

so the absolute value of C is

C = |C| = \sqrt{C_{x}^{2} + C_{y}^{2}} = \sqrt{10.7^{2} + 3.9^{2}} = 11.4

d) Correct answer is C:

A · B = A_{x}B_{x} + A_{y}B_{y} = (−6.1) · (−9.8) + (−5.8) · 4.6 = 33.1

Exercise 2 To good approximation, we may regard the oxygen molecules as point shaped bodies since the distance between them (300 Å) is much larger than the linear extent of each molecule (on the order 1 - 2 Å). The mass of an oxygen molecule is m(O_{2}) = (32 g/mol) / (6.02·10^{23} molecules/mol) = 5.32 ·10^{−23} g/molecule = 5.32 ·10^{−26} kg. So the gravitational force between the two oxygen molecules is

F_{g} = G\frac{m(O_{2})^{2}}{r^{2}} = 6.67 · 10^{−11} \frac{m^{3}}{kg · s^{2}} · \frac{(5.32 · 10^{−26})^{2} kg^{2}}{(300 · 10^{−10})^{2} m^{2}} = 2.09 · 10^{−46} N

Gravitational forces are always attractive. With an extra electron, each ion O_{2}^{−} has a charge q = −e. The electric force F_{e} between the two ions is therefore

F_{e} = \frac{1}{4πε_{0}} \frac{e^{2}}{r^{2}} = 9 · 10^{9} · \frac{(1.6 · 10^{−19})^{2}}{(300 · 10^{−10})^{2}} = 2.56 · 10^{−13} N

With the charges in units C (coulomb) and distance in m (meters), and moreover the SI value 9 · 10^{9} for the constant factor 1/4πε_{0}, the force must come out in the unit N (newton). The electric force between two charges of equal sign is repulsive. The ratio between the two forces is

\frac{F_{e}}{F_{g}} = \frac{2.56 · 10^{−13}}{2.09 · 10^{−46}} ∼ 10^{33}

This implies that gravitational forces between (”not too large”) charged bodies usually may be neglected in comparison with the electric force.

Exercise 3

The third charge is supposed to be equally far from the two others, on the positive y axis. Pytagoras then yields a position y = \sqrt{(\sqrt{5}a)^{2} − a^{2}} = 2a for the third charge.

The force between −q and q is attractive since they are of opposite sign. In absolute value, the two forces F_{1} and F_{2} must be equal:

F_{1} = F_{2} = \frac{1}{4πε_{0}} \frac{q^{2}}{(\sqrt{5}a)^{2}} = \frac{q^{2}}{20πε_{0}a^{2}}

From the figure, we see that F_{1} and F_{2} have equal y components but opposite x components. The resulting force is therefore

F = (F_{1y} + F_{2y}) ŷ
= 2F_{1y} ŷ
= −2F_{1} \frac{2}{\sqrt{5}} ŷ
= −\frac{q^{2}}{5\sqrt{5}πε_{0}a^{2}} ŷ

Here, ŷ denotes the unit vector in the y direction. We have used that the ratio between F_{1y} and F_{1} must be the same as the ratio between the vertical distance 2a and the distance between q and −q, i.e., \sqrt{5} a. Or: In order to find the y component of F_{1}, we must multiply with cosine of the angle α (see figure), which is 2/\sqrt{5}. With q = 2 · 10^{−6} C and a = 4 · 10^{−2} m we have

F = \frac{(2 · 10^{−6})^{2}C^{2}}{5\sqrt{5}π · 8.85 · 10^{−12}C^{2}/Nm^{2} · (4 · 10^{−2})^{2}m^{2}} ≃ 8.0 N

If q in x = −a is replaced by −q, one attractive and one repulsive force act on the third charge:

We see from the figure that the total force F′ points in the positive x direction. The x component of each of the partial forces is found by multiplying with cos β = 1/\sqrt{5}. Hence we see that F′ must be half as big as F, i.e.,

F′ = \frac{q^{2}}{10\sqrt{5}πε_{0}a^{2}} x̂ = (4.0 N)x̂

This exercise illustrates several basic elements:

• Electric charges influence each other (”interact”) with forces according to Coulomb’s law.

• Forces, whether they are electric or of some other kind, are vectors.

• For electric forces (and also for other forces), the superposition principle applies: The total force on a charge is found by adding the partial forces acting on it. Do remember that this is addition of vectors.

Exercise 4

a) Because of symmetry, it should be pretty obvious that the test charge Q is affected by zero net force, since the forces from pairs of charges cancel each other:

b) We remove one of the charges, e.g. the ”upper” one:

It is then immediately clear that the net force on Q equals the force due to the removed charge, but with opposite sign, i.e., -F_{2}. Here, we have used the superposition principle. Using mathematics, we could express the solution like this: Let ∑_{(6)} F_{i} denote the net force with all 6 charges present, and ∑_{(5)} F_{i} the net force after we have removed the charge that acted on Q with the force F_{2}. Then

∑_{(5)} F_{i} = ∑_{(6)} F_{i} − F_{2} = 0 − F_{2} = −F_{2}

c) Also with an odd number of q charges, e.g. 7, the net force on the test charge Q in the centre must be zero. Suppose the net force was not zero. A rotation of the whole system through an angle of 360/7 ^{∘} would then result in a net force with a different direction. However, the system has not been changed as a result of the rotation, so the force on Q cannot have changed either. Hence, it must be zero. If you are not convinced by this argument, you may simply calculate the net force. Put Q in the origin and one q on the x axis:

The net force on Q then becomes:

F = F_{x}x̂ + F_{y}ŷ
= F_{0} (1 + 2 cos 2π/7 + 2 cos 4π/7 + 2 cos 6π/7) x̂ +

F_{0} (0 + sin 2π/7 + sin(−2π/7) + sin 4π/7 + sin(−4π/7) + sin 6π/7 + sin(−6π/7)) ŷ

= F_{0} (1 + 1.247 − 0.445 − 1.802) x̂
= 0

Here, F_{0} denotes the magnitude of the force between q and Q, and we have used cos(−x) = cos x and sin(−x) = − sin x.

Exercise 5

a) The point charge q is in equilibrium if there is no force acting on it. Here, two forces are acting, one repulsive due to the charge 3q and one attractive due to −q. These two may cancel each other, but only if they point in opposite directions. This is not possible if q is located off the x axis, e.g. as in the figure above. (Here, dashed arrows denote partial forces and solid arrows the total force.) Hence, any equilibrium position must be on the x axis. b) We have been told that there is one equilibrium position x_{0} for q on the x axis. This cannot be between 0 and a, because on this interval, the two partial forces point in the same direction. Moreover, x_{0} cannot be to the left of x = 0: Here, the distance between q and 3q would always be shorter than the distance between q and −q. Hence, the repulsive force 3q^{2}/4πε_{0}x_{0}^{2} is always larger than the attractive force q^{2}/4πε_{0}(a − x_{0})^{2}. In conclusion, we must have x_{0} > a. The equilibrium position is found by setting the total force equal to zero:

0 = \frac{3q^{2}}{4πε_{0}x_{0}^{2}} x̂ − \frac{q^{2}}{4πε_{0}(x_{0} − a)^{2}} x̂
⇒ \frac{3}{x_{0}^{2}} = \frac{1}{(x_{0} − a)^{2}}
⇒ 3x_{0}^{2} − 6ax_{0} + 3a^{2} = x_{0}^{2}
⇒ 2x_{0}^{2} − 6ax_{0} + 3a^{2} = 0
⇒ x_{0} = \frac{6a}{4} + \frac{1}{4}\sqrt{36a^{2} − 24a^{2}} = \frac{3 + \sqrt{3}}{2}a ≃ 2.37a

The preassumption was x_{0} > a, so the solution with negative sign in front of the square root is not relevant. (It corresponds to x ≃ 0.63a, where both contributions to the total force are equal, and with the same direction.) The stability of the equilibrium position x_{0} is most easily found by looking at the net force when x ≫ x_{0}. Then the point charge q ”sees” approximately a point charge 3q − q = 2q and must experience a net repulsive force. We know that the force is zero only in x = x_{0}. This means that the force is pointing in the positive x direction for all x > x_{0}, and also for a small displacement to the right of x_{0}. Certainly, the force must be to the left if q comes fairly close to −q. Therefore, the force must be towards left also for a small displacement in the negative x direction. Alternatively, with a little bit of calculation: Let us first simplify notation by introducing the function f(x):

F (x) = F (x)x̂ = \frac{q^{2}}{4πε_{0}} (\frac{3}{x^{2}} − \frac{1}{(x − a)^{2}} ) x̂ ≡ \frac{q^{2}}{4πε_{0}} f(x)x̂

Next, we determine df/dx in x = x_{0}:

(\frac{df}{dx})_{x=x_{0}} = −\frac{6}{x_{0}^{3}} + \frac{2}{(x_{0} − a)^{3}} ≃ −\frac{6}{(2.37a)^{3}} + \frac{2}{(1.37a)^{3}} ≃ \frac{0.33}{a^{3}} > 0

Since f(x_{0}) = 0 and f′(x_{0}) > 0, the equilibrium is unstable.