Solutions to Homework Questions 2 Chapt16, Problem-1: A proton moves 2.00 cm parallel to a uniform electric field with E = 200 N/C. (a) How much work is done by the field on the proton? (b) What change occurs in the potential energy of the proton? (c) Through what potential difference did the proton move? (a) The work done is W = F!s cos" = qe ( )!s cos", or W = ( 1.60!10 "19 C) ( 200 N C) ( 2.00!10 "2 m )cos 0 = 6.40!10 "19 J (b) The change in the electrical potential energy is!pe e = "W =! 6.40" 10!19 J (c) The change in the electrical potential is!v =!PE e = "6.40 10 "19 J q 1.60 10-19 C =! 4.00 V Chapt16, Problem-3: A potential difference of 90 mv exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na + ) from the interior of the cell? The work done by the agent moving the charge out of the cell is ( ) W input =!W field =!!"PE ( e )=+q "V = 1.60!10 "19 ( C) + 90! 10 "3 J & ( = C 1.4!10 "20 J Chapt16, Problem-7: A pair of oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate? E =!V (a) d = 600 J C 5.33 "10 3 m = 1.13! 105 NC (b) F = q E= ( 1.60!10 "19 C) ( 1.13!10 5 NC)= 1.80! 10"14 N (c) W = F!s cos" = ( 1.80!10 "14 N) [( 5.33 " 2.90)!10 "3 m]cos 0 = 4.38!10 "17 J 1
Chapt16, Problem-8: Suppose an electron is released from rest in a uniform electric field whose strength is 5.90x10 3 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm? (a)!v = Ed = ( 5.90 "10 3 Vm)1.00 ( " 10 2 m )= 59.0 V (b) so v = 2 q!v = m e 1 2 m ev 2! 0 = "KE = W =!"PE e = q ("V ), ( ) ( )( 59.0 J C) 2 1.60 "10 19 C 9.11 "10 31 kg = 4.55! 106 ms Chapt16, Problem-15: Two point charges Q 1 = +5.00 nc and Q 2 = 3.00 nc are separated by 35.0 cm. (a) What is the electric potential at a point midway between the charges? (b) What is the potential energy of the pair of charges? What is the significance of the algebraic sign of your answer? k e q V =! i (a) i r i = 8.99! 10 9 N "m2 & 5.00!10 )9 C C 2 ( 0.175 m ) 3.00!10 )9 C& ( = 0.175 m 103 V (b) PE = k e q i q 2 r 12 = 8.99! 10 9 N "m2 & C 2 ( ( 5.00! 10 )9 C) ) 3.00! 10 )9 C 0.350 m ( ) =! 3.85 "10!7 J The negative sign means that positive work must be done to separate the charges (i.e., bring them up to a state of zero potential energy). Chapt16, Problem-19: In Rutherford s famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2e and masses of 6.64x10 27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.00x10 7 m/s directly toward the gold nucleus as in Figure P16.19. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. From conservation of energy, ( KE + PE e ) f = ( KE+ PE e ) i, which gives ( )( 2e) 0 + k e Qq = 1 r f 2 m! v 2 i + 0 or r f = 2 k eqq 2 m! v = 2 k e 79e 2 i m! v i 2 8.99! 10 9 N" m 2 & C 2 ( ( 158 ) ( 1.60! 10)19 C ) 2 r f = 6.64!10 )27 kg ( )( 2.00!10 7 ms) 2 = 2.74!10 "14 m 2
Chapt16, Problem-20: Starting with the definition of work, prove that at every point on an equipotential surface the surface must be perpendicular to the local electric field By definition, the work required to move a charge from one point to any other point on an equipotential surface is zero. From the definition of work, W = ( Fcos!)" s, the work is zero only if s = 0 or Fcos! = 0. The displacement s cannot be assumed to be zero in all cases. Thus, one must require that Fcos! = 0. The force F is given by F = qe and neither the charge q nor the field strength E can be assumed to be zero in all cases. Therefore, the only way the work can be zero in all cases is if cos! = 0. But if cos! = 0, then! = 90 or the force (and hence the electric field) must be perpendicular to the displacement s (which is tangent to the surface). That is, the field must be perpendicular to the equipotential surface at all points on that surface. Chapt16, Problem-25: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm 2, separated by a distance of 1.80 mm. If a 20.0-V potential difference is applied to these plates, calculate (a) the electric field between the plates, (b) the capacitance, and (c) the charge on each plate. E =!V (a) d = 20.0 V 1.80" 10-3 m = 1.11" 104 Vm= 11.1 kv m directed toward the negative plate ( )( 7.60" 10 4 m 2 ) (b) C =! 0 A = 8.85 " 1012 C 2 N m 2 d 1.80 "10-3 m = 3.74!10 "12 F = 3.74 pf (c) Q = C (!V )= 3.74" 10 12 F ( )= 7.47 "10 11 C = 74.7 pc on one plate and! 74.7 pc on the other plate. ( ) 20.0 V Chapt16, Problem-28: A small object with a mass of 350 mg carries a charge of 30.0 nc and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 4.00 cm. If the thread makes an angle of 15.0 with the vertical, what is the potential difference between the plates? mg!f y = 0 " T cos15.0 =mg or T = cos15.0!f x = 0 " qe = T sin15.0 =mg tan15.0 T 15.0 or mg tan15.0 E = q F = qe mgd tan15.0!v = Ed = q mg ( )( 9.80 m s 2 ) 0.0400 m!v = 350 "10 6 kg ( )tan15.0 30.0 " 10 9 = 1.23 "10 3 V = C 1.23 kv 3
Chapt16, Problem-31: (a) Find the equivalent capacitance of the group of capacitors in Figure P16.31. (b) Find the charge on and the potential difference across each. Using the rules for combining capacitors in series and in parallel, the circuit is reduced in steps as shown below. The equivalent capacitor is shown to be a 2.00 µf capacitor. 4.00 µf 3.00 µf 6.00 µf 3.00 µf 2.00 µf a b c a b c a c 2.00 µf 12.0 V Figure 1 (b) From Figure 3: Q ac = C ac!v 12.0 V Figure 2 ( ) 12.0 V ( ) ac = 2.00 µf 12.0 V Figure 3 ( )= 24.0 µc From Figure 2: Q ab =Q bc = Q ac = 24.0 µc Thus, the charge on the 3.00 µf capacitor is Q 3 = 24.0 µc Continuing to use Figure 2, (!V) ab = Q ab 24.0 µc = C ab 6.00 µf = 4.00 V, and (!V) 3 = (!V) bc = Q bc 24.0 µc = C bc 3.00 µf = 8.00 V From Figure 1, (!V) 4 = (!V) 2 = (!V) ab = 4.00 V and Q 4 = C 4!V Q 2 = C 2!V ( ) 4 = ( 4.00 µf) 4.00 V ( ) 2 = 2.00 µf ( ) 4.00 V ( )= 16.0 µc ( )= 8.00 µc Chapt16, Problem-44: Two capacitors C 1 = 25.0 µf and C 2 = 5.00 µf are connected in parallel and charged with a 100-V power supply. (a) Calculate the total energy stored in the two capacitors. (b) What potential difference would be required across the same two capacitors connected in series in order that the combination store the same energy as in (a)? (a) When connected in parallel, the energy stored is W = 1 2 C 1 (!V ) 2 + 1 2 C 2 (!V ) 2 = 1 ( 2 C 1 + C 2 )(!V) 2 [ ] 100 V = 1 25.0 + 5.00 2 ( )!10 "6 F (b) When connected in series, the equivalent capacitance is 1 C eq = 25.0 + 1 1! & µf = 4.17 µf " 5.00 From W = 1 2 C eq!v!v = 2W C eq = ( ) 2 = 0.150 J ( ) 2, the potential difference required to store the same energy as in part (a) above is 2( 0.150 J) 4.17 "10 6 F = 268 V 4
Chapt16, Problem-51: A model of a red blood cell portrays the cell as a spherical capacitor a positively charged liquid sphere of surface area A, separated by a membrane of thickness t from the surrounding negatively charged fluid. Tiny electrodes introduced into the interior of the cell show a potential difference of 100 mv across the membrane. The membrane s thickness is estimated to be 100 nm and its dielectric constant to be 5.00. (a) If an average red blood cell has a mass of 1.00x10 12 kg, estimate the volume of the cell and thus find its surface area. The density of blood is 1100 kg/m 3. (b) Estimate the capacitance of the cell. (c) Calculate the charge on the surface of the membrane. How many electronic charges does this represent? (a) V = m! = 1.00 "10 12 kg 1100 kg m 3 = 9.09! 10"16 m 3 Since V = 4!r3 3 (b) " 3V, the radius is r = 4! & A = 4!r 2 " = 4! 3V 4! & C =! " 0 A d = 5.00 23 ( ) 8.85!10 "12 C 2 N m 2 13, and the surface area is ( ) " = 4! 3 9.09 ( 10)16 m 3 4! & ( )( 4.54! 10 "10 m 2 ) 23 = 4.54! 10"10 m 2 100! 10 "9 = m 2.01! 10"13 F (c) Q = C (!V)= ( 2.01"10 13 F) ( 100 "10-3 V)= 2.01! 10 "14 C, and the number of electronic charges is n = Q e = 2.01!10 "14 C 1.60! 10-19 = 1.26! 105 C Chapt16, Problem-54: Charges of equal magnitude 1.00x10 15 C and opposite sign are distributed over the inner and outer surfaces of the cell wall in Figure P16.54. Find the force on the potassium on (K + ) if the ion is (a) 2.70 µm from the center of the cell, (b) 2.92 µm from the center, and (c) 4.00 µm from the center. (a) At r = 2.70 µm from the cell center, the ion is located inside the spherically symmetric charge distributions on the surfaces of the cell wall. From Gauss s law, the field is zero at such a location, so F = qe = q 0 ( )= 0. 5
(b) With spherically symmetric charge distributions, consider a spherical gaussian surface of radius r = 2.92 µm centered on the cell center. From Gauss s law, the electric field at points on the gaussian surface is the same as if the total charge inside the surface, Q =!1.00" 10!15 C, was located at the center. Thus,! F = qe = q k eq " r 2 & * = ( +1.60 10 (19! C) 8.99 10 9 N )m 2 ((1.00 10 (15 C) -, / " C 2 & ( 2.92 10-6 m) 2 +,./ F =!1.69" 10!13 N or F = 1.69! 10"13 N directed radially inward (c) At r = 4.00 µm from the cell center, the ion is located outside a spherically symmetric charge distribution having zero net charge. From Gauss s law, the field is zero at such a location, so F = qe = q 0 ( )= 0. Chapt16, Problem-55: A virus rests on the bottom plate of oppositely charged parallel plates in the vacuum chamber of an electron microscope. The electric field strength between the plates is 2.00x10 5 N/C, and the bottom plate is negative. If the virus has a mass of 1.00x10 15 kg and suddenly acquires a charge of 1.60x10 19 C, what are its velocity and position 75.0 ms later? Do not disregard gravity. The electric field is directed vertically downward with E y =!2.00 " 10 5 NC. Thus, a y =!F y m = qe y " mg m or a y =+22.2 m s 2 At t = 75.0 ms = 0.0750 s, v y = v yi + a y t = 0 + 22.2 m s 2 ( )("2.0010 5 NC) = "1.60 10 "19 C 1.00 10-15 kg ( ) 0.0750 s ( 22.2 m s2) 0.0750 s " 9.80 m s 2, ( )= 1.67 m s upward, and ( ) 2 y = v yi t+ 1 2 a y t 2 = 0 + 1 2 = 0.0624 m = 6.24 cm above bottom plate Chapt16, Conceptual-5: Suppose you are sitting in a car and a 20 kv power line drops across the car. Should you stay in the car or get out? The power line potential is 20 kv compared to the potential of the ground. [Adapted from the solution in the text] The rubber (a good insulator) tires on the car means that there is not good electrical contact between the rest of the car (including occupants) and the Earth (ground). Thus if the power line makes electrical contact with the metal of the car, it will raise the potential of the car to 20kV. It will also most-likely raise the potential of your body to 20kV, because you are undoubtedly in good electrical contact with the car. In itself, this is not a problem electrostatic equilibrium will be established, and charge will quickly cease to flow. However if you step out of the car, your body (with a potential of 20 kv) will make contact with the ground, which is at a potential of zero Volts (by definition) As a result, a current will pass through your body, and you are likely to be injured. Thus it is best to stay in the car until help arrives. 6
Chapt16, Conceptual-6: Why is it important to avoid sharp edges or points on conductors used in high-voltage equipment? A sharp point on a charged conductor would produce a large electric field in the region near the point (see Section 15.6). Therefore an unintended electric discharge could take place at such a location. Chapt16, Conceptual-9: [Sorry, I did not spot the typo in the textbook here is a rewording of the question] Why is it dangerous to touch the terminals of a highvoltage capacitor even after the voltage source that charged the battery capacitor is disconnected? from the capacitor? What can be done to make the capacitor safe to handle after the voltage source has been removed? The capacitor often remains charged long after the voltage source is disconnected. This residual charge can be lethal. The capacitor can be safely handled after discharging the plates by short-circuiting the device with a conductor, such as a screwdriver with an insulating handle. Chapt16, Conceptual-15: If you were asked to design a capacitor where small size and large capacitance were required. What factors would be important in your design? You should use a dielectric-filled capacitor whose dielectric constant is very large. Furthermore, you should make the dielectric as thin as possible, keeping in mind that it cannot be too thin or else dielectric breakdown will occur. 7