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Brock University Physics 1P22/1P92 Winter 2014 Dr. D’Agostino Solutions for Tutorial 5: Electric Fields and Forces (Chapter 20) 1. When I take clothes out of a clothes-dryer right after it stops, some of the clothes stick to me. Is my body charged? If so, how did it get a charge? If not, then why does this happen? [2 points] Solution: My body is almost certainly neutral. However, all the rubbing of the clothes with each other and with the metal drum of the dryer transfers charge so that some of the clothes end up being charged. When one of the charged pieces of clothing touches my uncharged body, my body becomes polarized, and then there is an attraction between the charged clothing and my uncharged body. The same effect is operative when you rub a balloon on your hair and then it sticks to a wall. The same effect also allows pollen to stick to bees, as explained in the textbook. 2. A light-weight neutral metal ball hangs by a thread. When a charged metal rod is held near, the ball moves towards the rod, touches the rod, and then quickly “flies away” from the rod. Explain each step in this behaviour. [2 points] Solution: When the charged rod is brought near the ball, charge polarization takes place in the ball, resulting in attraction between the ball and the rod. When the ball touches the rod, charge is transferred between the rod and the ball so that they each end up with a net charge of the same sign. Because the rod and ball have net charges of the same sign, they repell each other. 3. A typical commercial aircraft is struck by lightning about once per year. When this happens, the external metal skin of the airplane might be burned, but the people and equipment inside the aircraft experience no ill effects. Explain why this is so. [2 points] Solution: When lightning strikes, there is a tremendous transfer of charge from the cloud to the object struck. When lightning strikes the metal plane, this charge is distributed over the surface of the plane. There will initially be movement of charges over the surface of the plane (over a very short time interval) but a situation of static equilibrium will quickly be established. Since we have established that there is no electric field inside a conductor, the passengers and equipment inside the plane will experience no effect. Similarly, if you are inside a car struck by lightning, you will be safe. The tires will most likely be damaged as this tremendous amount of charge moves through them to ground. 4. An electric dipole is formed from ±1 nC point charges spaced 2.0 mm apart. The dipole is centred at the origin, oriented along the y-axis. Determine the strength and direction of the electric field at the point (10 mm, 0 mm). [4 points] Solution: Assume that the positive charge is at (0, 1.0 mm) and the negative charge is at (0, −1.0 mm). Of course, you could also make the opposite assumption, because the statement of the problem doesn’t specify this; if you make the opposite assumption, the electric field in the result will have the same magnitude but opposite direction. Draw a diagram! In your diagram, you’ll notice a key angle, 1 −1 = 5.71◦ θ = tan 10 The distance r from each of the charges to the point (10 mm, 0 mm) at which we wish to calculate the field satisfies 2 2 r2 = 1 × 10−3 + 10 × 10−3 = 1.01 × 10−4 m2 The electric field created by the positive charge at the point (10 mm, 0 mm) is KQ KQ KQ ~ E1 = cos θ, − 2 sin θ = 2 (cos θ, − sin θ) 2 r r r The electric field created by the negative charge at the point (10 mm, 0 mm) is KQ KQ KQ ~2 = − E cos θ, − sin θ = (− cos θ, − sin θ) r2 r2 r2 ~1 + E ~ 2: By the principle of superposition, the total electric field is the vector sum E ~ =E ~1 + E ~2 E ~ = KQ (cos θ, − sin θ) + KQ (− cos θ, − sin θ) E r2 r2 ~ = KQ (cos θ − cos θ, − sin θ − sin θ) E r2 ~ = KQ (0, −2 sin θ) E r2 The magnitude of the electric field is KQ (2 sin θ) r2 (8.99 × 109 ) (1 × 10−9 ) E= (0.199) 1.01 × 10−4 E = 18 kV/m E= The electric field is in the −y direction. Remember that if you placed the dipole in the opposite direction to the one chosen here then the magnitude of the electric field will be the same as calculated here, but the direction of the electric field will be in the +y-direction.