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by: Vernice Schuster

FundamentalsofPhysicsII PHYS102

Marketplace > Drexel University > Physics 2 > PHYS102 > FundamentalsofPhysicsII
Vernice Schuster
GPA 3.54


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This 3 page Class Notes was uploaded by Vernice Schuster on Wednesday September 23, 2015. The Class Notes belongs to PHYS102 at Drexel University taught by WilliamKing in Fall. Since its upload, it has received 24 views. For similar materials see /class/212515/phys102-drexel-university in Physics 2 at Drexel University.

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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


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