Physics 101 Homework Set 9 Solutions Chapter 22: Project 2: Yes, the stream of water will deflect (if it is nice and dry out and you can get a large charge on the comb or balloon). This is because the comb is charged and the water is neutral. The charge on the comb polarizes the water and attracts it. ------------------ Exercises: 4: Friction between the bag and the wool suit transfers electrons from one to the other. The bag becomes positively charged and the suit becomes negatively charged. 6: Friction between the car and the air and between the car and the road can transfer charge to the car. Since the rubber tires are insulators, the charge stays on the car until something touches the car. If the 'something' is a toll collector, he will get a nasty shock. To prevent this, the thin metal wire is positioned so that it will touch the car and transfer the excess charge on the car to the ground. 8: When the metal ball is touched by a charged body, it transfers charge to the ball. The charge spreads out, ie: some of it travels from the ball to the leaves. This means that there is excess charge on the leaves and that both leaves have the same sign charge. Since like charges repel, the two leaves now repel each other. You can make a simple electroscope in the classroom with very thin conducting foil, a jar, a stopper, and a paper clip. 10: No, it is not necessary for the charged body to touch the electroscope for the leaves to spread apart. This is because when you bring the charged body close to the electroscope, it polarizes the electroscope. If the charged body is negative, then it will repel negative charges, leaving the ball positively charged and the leaves positively charged. The leaves will then repel each other. 18: The 5000 billion billion electrons in a penny do not fly out of the penny because their mutual repulsion is balanced by their attraction to the 5000 billion billion protons in the same penny. Note that if there are fewer protons than electrons, the excess electrons will leave the penny as soon as they can. 20: If you reduce the distance to 1/2, then the force is 4 times larger. If the distance is 1/4, then force is 16 times larger. If the distance is four times larger, then the force is 16 times smaller. In all these cases, you use Coulomb's Law: F = k q1 q2 / d^2 If d is replaced by d(new) = (1/4)d, then F(new) = k q1 q2 / d(new)^2 = k q1 q2 / [(1/4)d]^2 = k q1 q2 / [ (1/16)[d]^2 ] and since 1 / (1/16) = 16 = 16 * k q1 q2 / d^2 = 16 F 24: If d doubles, then the Electric field decreases by a factor of 4. This is guided by Coulomb's Law where E = k q / d^2 (Note that the electric field equation has the same form as the electric force equation but there is only one charge involved.) 32: There is a mist of paint surrounding a neutral car. Initially, the paint and the car are neutral. Note that the mist consists of many tiny paint droplets separated by air (which is an insulator). We only need to analyze the behavior of one paint droplet (since all of them will behave the same way). When the car is charged, it will polarize the paint droplet. The charged car will then attract the neutral paint droplet. 38: If I do 10 J of work to push 1 C of charge into an electric field, then it will have 10 J of electric potential energy. This means that the electric potential (NOT potential _energy_) is electric potential = electric potential energy ------------------------- charge = 10 J / 1 C = 10 V When it is released, the potential energy converts to kinetic energy (just like gravitational potential energy converts to kinetic energy when you drop something). -------------------------------- Problem 2: Use Coulomb's law: F = k q1 q2 / d^2 We know F = 20 N, d = 6 cm = 0.06 m, and q1 = q2. We can multiply both sides by d^2 to get: F d^2 = k q1 q2 and we can divide both sides by k: F d^2 / k = q1 q2 and we know that q2 = q1 so F d^2 / k = q1^2 so q1^2 = 20 N * (0.06 m)^2 / 9*10^9 Nm^2/C^2 = 8 * 10^-12 C^2 q1 = sqrt(8 * 10^-12 C^2) q1 = 2.8 * 10^(-6) C --------------------------------- Estimation: 10^25 electrons = 10^25 e * (1.6 * 10^(-19) C/e) = - 1.6 * 10^6 C The minus sign comes because electrons are negative. Similarly 10^25 protons = +1.6*10^6 C since protons have a positive charge of exactly the same magnitude as the electron. The distance between them is twice the radius of the Earth = 2*6400 km = 1.3*10^7 m. Then F = 9*10^9 Nm^2/C^2 * (-1.6*10^6 C) * (+1.6*10^6 C) ----------------------------------------------- (1.3*10^7 m)^2 F = 1.4 * 10^8 N Since 1 lb = 4.4 N, this is 3.1 * 10^7 lb = 30 million pounds. Rather a lot! This is almost as much as the USS Wisconsin.