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Date Created: 09/23/15
Recitation 1 Chapter 19 Problem 3 Nobel laureate Richard Feynman once said that if two persons stood at arms length from each other and each person had p 1 more electrons than protons the force of repulsion between them would be enough to lift a weight equal to that of the entire Earth Carry out an order of magnitude calculation to substantiate this assertion Let m 70 kg be the mass of one person and 48 be the charge of one electron Assume that there are approximately equal numbers of protons electrons and neutrons in a person Electrons have much less mass than protons or neutrons so we ignore their mass contribution Protons and neutrons have very similar masses so N Tn2Tnp is the number of protons and Nq N p is the number of extra electrons in each person Assume they are seperated by T 1 m The force of repulsion F is given by 2 2 719 2 719 2 7 i 7 mpqe 7 I g I 2 2 70 kg 001 16 10 C N I 10 03 10 7 I 25 F 7 k8 T2 7 k8 727211 7 90 10 N m C 21 10727 kg 1 m N1 10 7110727 N 71 10 N 1 And a weight the mass of the earth would be Fg Mg m 61024 kg 98 ms2 m 61025 N N F Problem 4 Two protons in an atomic nucleus are typically seperated by a distance of T 200 10 15 m The electric repulsion force F between the protons is huge but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart What is the magnitude of F F 39 1 i r 16010 lg C 2 7 if I 9 I 2 2 F7keT278i99 10 NmC 2 003910715m 2 gt 577 N 2 Problem 9 1n the Bohr theory of the hydrogen atom an electron moves in a circular orbit about a proton where the radius of the orbit is T 0529 10 10 m a Find the magnitude of the electric force each exerts on the other b If this force causes the centripetal acceleration of the electron what is the speed of the electron V a 2 2 719 iii 39939221601007 3973 F7keT2 7899 10 NmC 0 5293910710m 7822 10 N 3 b Using FC maC mv2T FT 82410 3 N 0529 10710 m 7 219106 4 v V m 91110731 kg 15 H Problem 11 In Figure P1911 determine the point other than in nity at which the electric eld is zero ql 7250 MC and qg 600 MC T1 7 1 1 11 42 E2 E1 X First we need a coordinate systemi Let ql be the origin 11 0 and qg be at 12 1100 m The electric eld of a nite number of point charge is given by p 612 1916 4139 h E ks 721 5 For any point off the z axis there would be some force moving the charge in the vertical y direction so we only need to look at positions on the z axis A positive test charge placed between the two charges would be pulled to the left by ql and pushed to the left by 421 A positive test charge placed to the right of qg would be pushed to the right by qg more strongly because qg gt ql and 7 2 lt T1 than it would be pulled to the left by 411 So the only place to look for equilibrium is to the left of 41 z lt 0 where 7 2 7 1 12 41 42 Eke iiii 0 6 Ti T1I22gt 42 41 i 7 7 Ti T1 I22 712 1 E i j 8 7 1 T1 41 I2 7 1 1182 m701392 m 9 4 i12 7 1 41 But 7 1 70392 m is between the two charges where our assumption about the electric elds opposing each other doesnlt hold so E 0 only at a 7 1 18 m z 71182 m Problem 15 Four point charges are at the corners of a square of side a as shown in Figure P1915 with ql 21 L12 Sq qg 4g and q4 q a Determine the magnitude and direction of the electric eld at the location of charge 441 b What is the resultant force on 44 Let point to the right and point up 7 hli 2i 34 4i 7 i 2 i c c E ker2 18 via 2 a 4 lt2 NEOHHA J 10 2 2 q 3 3 q Ekei yl 2 7 4 7 5191kei 11 a2 ltN gtltN gt a2 And the direction 19 measured counter clockwise from of E is given by 4 i 19 arctanlt 2 58180 12 2 27 So the magnitude of E is given by b F qE so the direction of F is the same as the direction of E1 The magnitude of F is given by F 531165412a2 Problem 19 A uniformly charged ring of radius T 1010 cm has a total charge of q 7510 0 Find the electric eld on the axis of the ring at a Ia 1100 cm b I 5100 cm c 10 3010 cm and d Id 100 cm from the center of the ring