Physics 9 Spring 2011 Homework 1 - Solutions Wednesday January 19, 2011

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1 Physics 9 Spring 011 Homework 1 - s Wednesday January 19, 011 Make sure your name is on your homework, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don t finish them. The homework is due at the beginning of class on Wednesday, January 6th. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours. 1. A.0 µc point charge and a 4.0 µc point charge are a distance L apart. Where should a third point charge be placed so that the third charge has no acceleration? The third charge should be placed at a point where the electric force from the other two charges cancels out. Suppose that the third charge is positive. Then it will be attracted to the.0 µc charge, and repelled by the 4.0 µc charge. However, since the positive charge is bigger we should place the third charge to the left of the negative charge since the repulsion will be smaller than the attraction. Suppose that the third charge is negative. In this case the forces will be opposite, but we still see that we should place the charge to the left of the.0 µc charge. So, suppose that we call the distance from the negative charge to the third charge x. Then, the attractive force on the third charge (denoted q) from the negative charge will be F = 1 ( 10 6 ) q. 4πɛ 0 x The repulsive force from the other charge will be F + = 1 4πɛ 0 ( ) q (x + L). If the third charge is to be static then the two forces must cancel, F = F +. The third charge cancels off (meaning it doesn t matter whether it s positive or negative), and we have 10 6 = x (x + L) (x + L) = x. Squaring and rearranging gives x xl L = 0 s = L ± 4L + 4L = L ± L. We want the third charge to be to the left of the negative charge, so we take the negative root. So, we need the third charge on the left of the µc charge, a distance ( x = 1 ) L 0.41L. 1

2 . A point particle that has charge +q and unknown mass m is released from rest in a region that has a uniform electric field E that is directed vertically downward. The particle hits the ground at a speed v = gh, where h is the initial height of the particle. Find m in terms of E, q, and h. When the particle starts out at height h, it has two forces acting on it, both pointing downward: gravity, F g = mg, and the electric force F E = qe. So, F net = qe + mg, and is constant. So, the constant acceleration is a = F net /m = g + qe. For a constant m acceleration, the particle travels a distance h in time t, such that h = 1 h at t = a. So, after this time, the particle is moving at velocity v = at = a = ha. But, we re told that it hits the ground with speed v = gh, and so gh = ha, or a = g. Since a = g + qe/m, we see that qe m = g m = qe g. (You could figure out the form of this expression just by looking at the units.) h a

3 3. An electron (charge e, mass m) and a positron (charge +e, mass m) revolve around their common center of mass under the influence of their attractive Coulomb force. Find the speed v of each particle in terms of e, m, k, and their separation distance L. Since the two charges have the same mass, they will have the same speed. If they are separated by a distance L then the electrical force between them is just the Coulomb force, F = ke L. However, the particles are moving around in a circle and so the net force on them is the centripetal force, with radius r = L/, F = mv L/ = mv L = ke L. Solving for the speed gives v = ke ml. 3

4 Applying Coulomb s law for electric fields and the superposition of fields in interval IV yields: 3.00 μc + ( x 0) ( x 0.00 m) ( x 0.30 m) 4. Two charged spheres, each of mass m and charge q, are suspended from a common point by strings interval of lengthiv. L. Each string makes an angle θ with the vertical as shown in the figure to the right. Summarizing, (a) Show thatthe electric field is zero q = L sin θ (mg/k) tan θ, where k is the Coulomb constant μc where k is the Coulomb constant. (b) Find q if m = 10.0 g, L = 50.0 cm, and (a) The two spheres are in static equilibrium, such that the sum of the forces on each is zero. There are three forces acting on them, as seen in the figure to the right. There is the gravitational force, pointing down, the tension pointing up along the string, and the Felectrical r force, pointing to the g, and the electric right (for the right-hand sphere). Working out F r the components, E. We can use Coulomb s law to Fx = T sin θ + F e = 0 Fy = T cos θ mg = 0. So, solving for tension in the first equation, and plugging in into the second tells us that 97. μc = The roots of this equation are at x = m and x = 0.0. Neither of these are in at two locations in interval II: x = m and x = m 74 Two point particles, each of mass m and charge q, are suspended from a common (b) Find point q if m by = threads 10.0 g, L of = length 50.0 cm, L. Each and thread makes an angle θ with the θ = q = Lsinθ mg k tan vertical as shown in Figure (a) Show that ( ) θ θ = 10.0º. Picture the Problem Each point particle is in static equilibrium under the influence of the tensiont r, the gravitational force force relate the electric force to the charge on each particle and their separation and the conditions for static equilibrium to relate these forces to the charge on each particle. F e = mg tan θ. Now, the electrical force is just the Coulomb law, F e = kq, where r is the separation distance between the charges. The separation distance can be written in r terms of the length of the string and angle as such that the forces equal to Solving for the charge gives r = L sin θ, kq 4L sin θ = mg tan θ. mg tan θ q = L sin θ, k 4 T r θ y 0 F r e r r F = mg g x

5 as claimed. (b) Here we just need to plug in the given numbers, mg tan θ q = L sin θ k = 0.5 sin (10 ) = 0.41 µc tan (10 ) 5

6 5. A point particle of mass m and charge q is constrained to move vertically inside a narrow, frictionless cylinder, as seen in the figure to the right. At the bottom of the cylinder is a point charge Q having the same sign as q. (a) Show that the top particle will be in equilibrium at a height y 0 = /mg. (b) Show that if the particle is displaced from its equilibrium postion by a small amount and released, it will exhibit simple harmonic motion with angular frequency ω 0 = g/y 0. Hint: use the binomial expansion (1 + ɛ) α 1 + αɛ, for small ɛ. (a) The equilibrium position, which we ll call y 0, occurs when the electrical force balances the gravitational force, F E = F g, or = mg y y0 0 = mg. (b) Now, suppose that we shift move the top charge by a small distance y away from the equilibrium position. Then it s no longer in equilibrium and so it experiences a net force, mg = ma, (y 0 + y) where a is the acceleration of the charge. Now, let s rewrite the first term using the binomial expansion, (1 + ɛ) α 1 + αɛ, for small ɛ, = 1 (y 0 +y) y0 (1+y/y 0 ) = y0 y0 (1 + yy0 ) (1 yy0 ) = mg mg y 0 y, where we have used our result from part (a) to rewrite y 0. Thus, the total force is ma = mg mg y mg = mg y. y 0 y 0 6

7 Finally, the acceleration may be written as a = d y dt, such that d y dt + g y = 0. y 0 This is precisely the differential equation of a harmonic oscillator, d y dt + ω 0y = 0, if we identify ω 0 = g/y 0. 7