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AP Physics B Electrostatics Review name comb
AP Physics B Electrostatics Review Name__________________ 1. Describe the similarities and differences between the laws of gravity (Newton's Universal Gravitation) and Coulomb's law of electrostatic force.(write equations) Similarities: both deal with forces between two objects and both have the inverse square law in them. Differences: Charge vs. mass and the constant is different 2. State FOUR ways an object can be charged and explain each. Friction – rubbing to transfer electrons Conduction – contact is involved. An object is charged through contact with a charged object. Induction – non-contact charging. Ground a neutral object and place a charged object nearby. Polarization – no charge removal, e- and p+ are polarized in the molecules 3. When a comb attracts a tiny piece of paper sometimes one piece jumps away from the comb. Explain why. The piece may be the same charge as the comb, thus would repel. 4. Draw the electric field lines surrounding two NEGATIVE charges a distance d apart. 5. Find the magnitude and direction of the force between two charges of 1.5 nanocoulombs (1 nc = 1 x 10-9 C) one millimeter apart. 2 kq q 9.0 109 Nm 2 / C 2 1.5 109 C Fe 12 2 0.02 N , away from each other r (0.001) 2 a. What would happen to this force if the charges were tripled and the distance between them were reduced to two thirds its original value? k 3q1 3q2 kq1q2 9 9 81 Fe 2 2 9 , Fe would increase 81/4 times. 2 r 4/9 4 4 r 3 1 6. What is the net force on an UNCHARGED conducting sphere in a UNIFORM electric field? zero If E = 0, then by definition, E = F/q. Therefore, F must be 0. 7. Two parallel conducting plates are connected to a voltage source. What happens to the magnitude of the electric field if the voltage is doubled and the distance between the plates is reduced by half? V 2V V V Ed , E , 4 , 4 times as much. d 1 d d 2 8. A small conducting sphere holds charge +Q. Another sphere holding charge -5Q is R centimeters away from the first sphere. The force between the charges is 25 Newtons. The charges are briefly brought into contact and then returned to their original positions. What is the force between the charges now? Has its direction changed? kq1q2 kQ5Q kQ 2 Originally, F = 25 N, Fe 2 , 25 , 5 2 r R2 R k 2Q 2 kQ 2 After coming together they are both now -2Q, so F 4 , R2 R2 And, Fnew = 4 x 5 = 20 N 9. What is true about the electric field inside a conductor? Why? E = 0, in a static situation, charges must remain at rest. Therefore, the net electric force on them must be zero. 10. What is true about the direction of an electric field just outside a conductor? It is perpendicular to the surface of the conductor. 11. A charge -Q is placed at the center of a spherical conducting shell. Draw the pattern of electric field lines. 2 12. Two very large parallel metal plates 5 cm apart are oppositely charged with a potential difference of 10,000 volts between them. As an electron is moved from the negative to the positive plate (a) What happens to the electric potential energy of the electron? Decreases. e- (b) What happens to the electrostatic force on the electron? Increases. (c) What happens to the electric potential at the location of the electron? Does not change. (d) Calculate the electric field strength midway between the plates. V 10, 000V E 400, 000 N / C d 0.025m (e) If the electron were released at the negative plate, what would be its speed when it reached the positive plate? U e q0 Ed q0V 1.6 1019 C 10,000V 1.6 1015 J U e KE 1/ 2mv 2 v 2U e 2 1.6 1015 J 5.93 10 m / s 7 m 9.1110 31 kg 13. Charges +Q and -Q are located a distance L meters apart as shown below: L +Q -Q P (a) Find an expression for the electric field strength (and its direction) at a point midway between the charges on the line connecting them. Skip right now. (b) Find an expression for the electric potential at point P. Skip right now. (c) Repeat part (b) with both charges equal to +Q. Skip right now. (d) Find an expression for the electric field strength at point P shown above. P is equidistant from the charges but lies a distance L/2 below them. What is the direction of the field? Skip right now. 3 14. Two particles of charge Q are located at opposite corners of a square of side d. (a) What is the direction of the electric field at each of the other corners? Q Q @ a 45 diagonal from corners (b) What is the potential energy of a particle of charge +q held at these other corners? kq q kQ2 U 1 2 2 2 r12 d 15. A 1 keV electron (mass 9.11 x 10-31 kg) enters the region between two parallel conducting plates as shown below. (note 1 eV = 1.6 x 10-19 Joules) In this region the electric field strength is 10,000N/C. The plates are 5.0 cm apart: + e- - (a) Find the potential difference across the plates. V Ed 10,000N / C 0.05m 500V (b) Which plate is at higher potential? Negative: e- going from – to + is easy. Therefore there is more potential at a negative plate where you must force the negative charge there. (c) Find the force on the electron. F Eq 10,000 N / C 1.6 1019 C 1.6 1015 N (d) Find the acceleration of the electron toward the positive plate. F 1.6 1015 N a 1.76 1015 m / s 2 m 9.111031 kg (e) Does the electron gain or lose energy as it moves between the plates? Explain. Moving perpendicular to the field, it will not gain or lose energy. Going from + to – plate, it will gain Going from – to + plate, it will lose 4 16. A charged pith ball of mass m = 1.2 grams is held by a string (to form a pendulum) in a region of uniform electric field E = 5000N/C horizontal and to the right. The charge on the pith ball is 1.5 x 10-7C. (a) Find the electrostatic force on the pith ball F Eq 5000 N / C 1.5 107 C 7.5 104 N (b) DRAW the situation. Be careful about which way the string hangs. E (c) Make a free body diagram on which you identify all the forces. FT Fe Fg (d) Find the angle from the vertical at which this "pendulum" hangs. Fg 10m / s 2 0.0012kg 0.012N Fe 7.5 104 N Fe F 7.5 104 N tan tan 1 e tan 1 3.6 from vertical Fg Fg 0.012 N (e) The string is cut. Describe the motion of the pith ball. It will fall almost straight down, but will make a slightly curved path to the right. The electric field will force it to curve to the right as it goes down. 5 Capacitance Questions + - Capacitors store electric charge on parallel conducting (metal) plates closely spaced together. Usually they are rolled up with wax paper in between metal foil layers. The amount of charge that can be stored is Q = CV, where the constant C (for that capacitor) is called capacitance (unit "farad"). For a parallel plate capacitor which has plates of area A separated by distance d the capacitance is C = o A/d. o is called the permittivity of free space. o = 8.85 x 10-12 C2/N m2. The energy stored on a capacitor is U = ½ QV = ½ CV2 = ½ Q2/C 18. A certain capacitor has a capacitance of 50 pf (picofarads: 1 pf = 1.0 x 10-12 farad) (a) If the plates are separated by 1.0 x 10-4 meters, what is their area? dC 110 m 50 10 F 4 12 A 5.6 104 m 2 0 12 2 8.85 10 C / Nm 2 (b) When connected to 5.0 volts, what charge can this capacitor hold? Q CV 50 1012 F 5V 2.5 1010 C (c) What energy can this capacitor store? 1 1 U QV 2.5 1010 C 5V 6.25 1010 J 2 2 (d) What charge could this capacitor hold when connected to 10 volts? Q CV 50 1012 F 10V 5.0 1010 C (e) What energy could it hold when connected to 10 volts? 1 1 U QV 2.5 1010 C 10V 2.5 109 J 2 2 (f) What would happen if the capacitor was connected to a much higher voltage, say 1000V? Q CV 50 1012 F 1000V 5.0 108 C It just has a bigger charge on the plates and the potential energy stored increases. 6