Physics 111 Fall 2007 Electrostatic Forces and the Electric Field  Solutions


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1 Physics 111 Fall 007 Electostatic Foces an the Electic Fiel  Solutions 1. Two point chages, 5 µc an 8 µc ae 1. m apat. Whee shoul a thi chage, equal to 5 µc, be place to make the electic fiel at the mipoint between the fist two chages equal to zeo? Fom the iagam below we see that the chage has to be to the ight of the 5µC chage locate at some istance. Measuing istances fom the leftmost chage, hee the 5µC chage, we ll place the aitional 5µC chage at a istance away. At the mipoint between the fist two chages the electic fiel has to vanish, so we 5 C 8 C 5 C have E net = 0 = k + iˆ. Solving fo we fin, = ( 0.6m) ( 0.6m) ( 0.6m) 0.97m fom the leftmost chage. 5µC 8µC 5µC x 0m 0.6m. The electic fiel insie biological membanes is extemely high, oughly 1 x 7 N/m. If this electic fiel geneate the only foce on a soium ion, what woul its acceleation be? Hee we nee the mass of a soium ion. The atomic mass of soium is amu, so assuming that the poton an neuton have the same mass (1.67x 7 kg) we have 3.67x 6 kg fo the mass of the soium ion. The chage on the ion is 1e  = 1.6x 19 C. Thus the acceleation 19 7 N q 1.6 C 1 fom Newton s n m 13 m law is a = E = = s m 3.67 kg 3. Fin the foce on a 5 µc point chage locate at a vetex on an equilateal tiangle of 0.5 m sies if µc point chages ae locate at the othe two vetices. Fom the iagam below an the symmety in the poblem, we see that the hoizontal components of the foce cancel while the vetical components of the foce a. Thus
2 y 5µC 60 o = 0.5m F F x y F net µc 60o µc x = 0N 1 9 Nm 5 C = k sin 60 = 9 C ( 0.5m) = 3.1Nj ˆ = 3.1N in the positive y iection C sin 60 = 3.1N 4. How close must two electons be if the electic foce between them is equal to the weight of eithe at the Eath s suface? Set the magnitue of the electic foce equal to the magnitue of the foce of gavity an solve fo the istance. e F = F k = E G 19 ( ) 9 ( N m C ) 31 ( 9.11 kg)( 9.80m s ) k = e = 1.60 C = 5.08m 7 5. A poton ( m = 1.67 kg) is suspene at est in a unifom electic fiel into account gavity at the Eath s suface, an etemine E. E. Take Since the gavity foce is ownwa, the electic foce must be upwa. Since the chage is positive, the electic fiel must also be upwa. Equate the magnitues of the two foces an solve fo the electic fiel. 7 ( 1.67 kg)( 9.80 m s ) ( 1.60 C) F = F qe= E= = = E G 19 q N C, up
3 6. In a simple moel of the hyogen atom, the electon evolves in a cicula obit 6 aoun the poton with a spee of 1.1 m s. Detemine the aius of the electon s obit. [Hint: what o you ecall about cicula motion?] The electic foce must be a aial foce in oe fo the electon to move in a cicula obit. mv F = F k = E aial obit obit 19 ( 1.60 C) ( ) 31 6 ( 9.11 kg)( 1.1 m s) 9 = k = = obit mv N m C.1 m 7. A small lea sphee is encase in insulating plastic an suspene vetically fom an ieal sping ( k = 16 N m) above a lab table, shown below. The total mass of the coate sphee is kg, an its cente lies 15.0 cm above the tabletop when in equilibium. The sphee is pulle own 5.00 cm below equilibium, an electic chage = 3.00 C is eposite on it an then it is elease. Using what you know about hamonic oscillation, wite an expession fo the electic fiel stength as a function of time that woul be measue at the point on the tabletop (P) iectly below the sphee. The sphee will oscillate sinusoially about the equilibium point, with an amplitue of 5.0 cm. The angula fequency of the sphee is given by ω = km= 16 N m kg = 1.5a s. The istance of the sphee fom the table is given by [ ( )] = cos 1.5t m. Use this istance an the chage to give the electic fiel value at the tabletop. That electic fiel will point upwas at all times, towas the negative sphee. ( 9 )( N m C 3.00 C ) 4.70 [ cos ( 1.5 )] m [ cos ( 1.5 )] E = k = = t t = 1.08 [ 3.00 cos ( 1.5t )] 7 N C, upwas NC
4 8. Given the two chages shown below, at what position(s) x is the electic fiel zeo? Is the fiel zeo at any othe points, not on the x axis? On the xaxis, the electic fiel can only be zeo at a location close to the smalle magnitue chage. Thus the fiel will neve be zeo to the left of the mipoint between the two chages. Also, in between the two chages, the fiel ue to both chages will point to the left, an so the total fiel cannot be zeo. Thus the only place on the xaxis whee the fiel can be zeo is to the ight of the negative chage, an so x must be positive. Calculate the fiel at point P an set it equal to zeo. ( ) E = k + k = x = ( x+ ) x= 1 ( + ) x x 0.41 The fiel cannot be zeo at any points off the xaxis. Fo any point off the xaxis, the electic fiels ue to the two chages will not be along the same line, an so they can neve combine to give Two point chages, + an of mass m, ae place on the ens of a massless o of length L, which is fixe to a table by a pin though its cente. If the appaatus is then subjecte to a unifom electic fiel E paallel to the table an pepenicula to the o, fin the net toque on the system of o plus chages. The electic fiel will put a foce of magnitue F = E E on each chage. The istance of each chage fom the pivot point is L, an so the toque cause by each EL foce is τ = F th toques will ten to E =. Bo make the o otate counteclockwise in the iagam, EL an so the net toque is τ = = EL net. F E E F E
5 . Fou equal positive point chages, each of chage 8.0 µ C, ae at the cones of a squae of sie 9. cm. What chage shoul be place at the cente of the squae so that all chages ae at equilibium? Is this a stable o unstable equilibium in the plane? A negative chage must be place at the cente of the squae. Let = 8.0 µ C be the chage at each cone, let q be the magnitue of negative chage in the cente, an let = 9.cm be the sie length of the squae. By the symmety of the poblem, if we make the net foce on one of the cone chages be zeo, the net foce on each othe cone chage will also be zeo. 1 F F 4 43 F 4 F 4q 41 q F = k F = k, F = x 41y o F = k F = k cos45 = k, F = k 4 4 x 4 y 4 4 F = k F = 0, F = k 43 43x 43 y q q q cos 45 o F = k F = k = k = F 4q 4qx 4qy 3 The net foce in each iection shoul be zeo. F = k + k + k = q = x + = 4 4 q C So the chage to be place is q = 7.66 C. This is an unstable equilibium. If the cente chage wee slightly isplace, say towas the ight, then it woul be close to the ight chages than the left, an woul be attacte moe to the ight. Likewise the positive chages on the ight sie of the squae woul be close to it an woul be attacte moe to it, moving fom thei cone positions. The system woul not have a tenency to etun to the symmetic shape, but athe woul have a tenency to move away fom it if istube.
6 11. A lage electoscope is mae with leaves that ae 78cmlong wies with tiny 4g sphees at the ens. When chage, nealy all the chage esies on the sphees. If the wies each make a 30 angle with the vetical as shown below, what total chage must have been applie to the electoscope? Ignoe the mass of the wies. The wies fom two sies of an equilateal tiangle, an so the two chages ae sepaate by a istance = 78 cm an ae iectly hoizontal fom each othe. Thus the electic foce on each chage is hoizontal. Fom the feeboy iagam fo one of the sphees, wite the net foce in both the hoizontal an vetical iections an solve fo the electic foce. Then wite the electic foce by Coulomb s law, an equate the two expessions fo the electic foce to fin the chage. F E F T θ F = F cosθ = 0 F = y T T cosθ F = F sinθ F = 0 F = F sinθ = sinθ = tanθ x T E E T cosθ ( ) tanθ F = k = tan θ = E k ( ) ( )( ) 9 ( N m C ) 3 o 4 kg 9.80 m s tan 30 = = m C 6.1 C
7 1. Suppose that electical attaction, athe than gavity, wee esponsible fo holing the Moon in obit aoun the Eath. If equal an opposite chages wee place on the Eath an the Moon, what shoul be the value of to maintain the pesent obit? 4 Use these ata: mass of Eath = 5.98 kg, mass of Moon = 7.35 kg, aius of obit = m. Teat the Eath an Moon as point paticles. Set the Coulomb electical foce equal to the Newtonian gavitational foce on one of the boies (the Moon). M M F = F k = G Moon Eath E G obit obit GM M 11 4 ( 6.67 N m kg )( 7.35 kg)( 5.98 kg) 9 ( N m C ) Moon Eath 13 = = = 5.71 C k
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