BROCK UNIVERSITY. Test 1 Solutions

Transcription

1 Examination date: 20 June 2013 Instructor: S. D Agostino BROCK UNIVERSITY Test 1 Solutions 1. A homemade capacitor is made of two parallel 35 cm by 35 cm sheets of aluminum foil separated by a 0.25 mm gap. (a) [2 marks] Determine the capacitance when the gap in the capacitor is empty. (b) [3 marks] Now the space between the two sheets of foil are filled with a dielectric of dielectric constant 2.4. Determine the charge on the positively-charged sheet if the potential difference between the plates is 10 V. Solution: (a) Because the capacitor is formed of parallel plates, the capacitance is C = ɛ 0A d C = C = 4.3 nf (b) The capacitance is now increased by a factor of 2.4, so C = κɛ 0A d C = κ C = nf C = 10 nf Thus, the charge on the positive plate is Q = C V Q = Q = 100 nc 2. [5 marks] The electric field strength is 50,000 V/m inside a parallel-plate capacitor with a 2.0 mm spacing. A proton is released from rest at the positive plate. Determine the proton s speed when it reaches the negative plate. Solution: Draw a diagram!

2 One way to solve the problem is to choose the potential of the negative plate to be 0 V. Then the potential of the positive plate is 50, 000 V/m 2.0 mm = 50, 000 V/m m = 100 V Now use conservation of mechanical energy, using subscript 1 to represent the positive plate and subscript 2 to represent the negative plate: U 2 + K 2 = U 1 + K 1 qv mv2 2 = qv mv2 1 q(0) mv2 2 = qv m(0)2 v2 2 = 2qV 1 m 2qV1 v 2 = m 2 ( ) (100) v 2 = v 2 = m/s 3. [4 marks] 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). 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 θ = tan 1 = 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 = m 2 The electric field created by the positive charge at the point (10 mm, 0 mm) is ( KQ E 1 = cos θ, KQ ) sin θ = KQ (cos θ, sin θ) The electric field created by the negative charge at the point (10 mm, 0 mm) is ( E 2 = KQ cos θ, KQ ) sin θ = KQ ( cos θ, sin θ)

3 By the principle of superposition, the total electric field is the vector sum E 1 + E 2 : E = E 1 + E 2 E = KQ (cos θ, sin θ) + KQ ( cos θ, sin θ) E = KQ (cos θ cos θ, sin θ sin θ) E = KQ (0, 2 sin θ) The magnitude of the electric field is E = KQ (2 sin θ) E = ( ) ( ) (0.199) E = 18 kv/m 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. 4. [6 marks] Write a concise paragraph to explain THREE ONLY of the following items. Remember, brevity and clarity are courtesy. NOTE: These paragraphs are samples only; your own words to similar effect will be good. (a) Give three every-day examples of charge transfer, explaining how the charge transfer occurs and the noticeable consequences. Solution: There are many every-day examples of charge transfer, of which here are a few: Brushing your hair on a dry winter day, charge being transferred between a road and a moving car, peeling plastic tape off a roll, and so on. In each case, it is friction between the two moving objects that transfers charge from one object to the other. The consequences in the first case is that individual strands of your hair repel each other, and therefore move away from each other ( standing on end ). In the second case, charge can build up on a car so that you might receive a shock if you touch the body of the car while standing on the ground. (The spark from the shock might ignite gasoline fumes if it occurs while you are fueling the car.) In the third case, the charge on the strip of tape may become noticeable if you try to tape down a very small paper object. There are many other examples. (b) Explain the field concept, contrasting it with instantaneous action at a distance. Solution: It s a bit mysterious how two charges could exert forces on each other when they are not in contact. However, with the field concept some of the mystery is removed (or at least transferred elsewhere). Each charge creates an electric

4 field in the space around it. What this electric field actually is remains somewhat mysterious; Faraday spoke of a tension in space, and nowadays we speak of a certain condition in space. In any case, the space around a charge has certain properties thanks to the electric field, and then another charge in this space experiences the effects of the field at its location. The fact that a charge experiences the effects of a field at its location takes some of the mystery out of action at a distance, and replaces it by a local effect. (c) Use an example to explain how it might be possible for two neutral objects to attract each other with an electric force. Solution: If one of the neutral objects is already polarized, then it can induce the opposite polarization in the other neutral object, after which the two objects will exert a net force on each other. A diagram will illustrate the details. (d) Explain how the electrostatic air filter on a forced-air gas furnace works. Solution: The strong electric field in the air filter either polarizes dust particles or charges them. Then the electric field forces the dust particles to one or the other of the plates, where the dust accumulates until the filter is removed and the dust is washed off. (e) Describe the properties satisfied by electric field lines and equipotential surfaces. Solution: Electric field lines start on positive charges and end on negative charges. The electric field is strongest where the field lines are closest together, and weakest where they are furthest apart. The direction of the electric field at each point along a field line is tangent to the field line. Equipotential surfaces are perpendicular to field lines where they intersect; thus, the direction of the electric field is perpendicular to an equipotential, pointing in the direction of decreasing potential. The electric field is strongest where equipotentials are closest together, and weakest where they are furthest apart. (f) Explain how electric field and electric potential are related. Solution: Equipotential surfaces are perpendicular to field lines where they intersect; thus, the direction of the electric field is perpendicular to an equipotential, pointing in the direction of decreasing potential. The electric field is strongest where equipotentials are closest together, and weakest where they are furthest apart. (g) Describe and explain the key facts about a conductor placed in a region of space containing an electric field. Solution: When the conductor reaches equilibrium, any excess charge is on the surface of the conductor the electric field inside the conductor is zero the exterior electric field is perpendicular to the surface of the conductor at each point on the surface of the conductor the external electric field is strongest near sharp points on the conductor the entire conductor is at the same potential

5 (h) Explain electric shielding, including a few examples from daily life. Solution: Because the total electric field inside a conductor (and inside a hollow within a conductor) is zero at equilibrium, an object placed within a hollow inside a conductor will be shielded from external fields. For example, sensitive electronic devices are often shipped inside containers that are lined with metal foil. Another example is coaxial cables, where the TV signal is shielded from interference from stray electric fields by a metal mesh covering the cable. Another example is the loss of radio signals when you drive under a metal bridge; the mesh of girders is almost as effective at blocking the radio signals as a solid metal object. (i) Explain a topic of your choice, taken from the material we ve studied so far in the course. (Make it good!)