Physics 260 Homework Solution 9 Chapter 23 1 a PSE6 23 P 002 47 electrons 6 02 10 23 atoms atom mole m numbers of electrons g 107 87 mole b number of electrons added Q 1 602 10 19 C Divide this value by 109 to get numbers of electrons added for every 109 electrons already present 2 PSE6 23 P 005 a It is a repulsive force since protons have positive charges The magnitude of the force is given by the Coulomb s law ke q 2 F 2 r b Gm1 m2 r2 FE ke q 2 FG Gm2 where G 6 673 10 11 Nm2 kg2 and ke 8 988 109 Nm2 C2 c ke q 2 Gm2 r2 r2 FG q m s 1 G ke 3 PSE6 23 P 008 The numbers of electrons and protons present are the same number of electrons protons 6 02 2023 electrons mole m g 1 mole The force between electrons and protons is then F k e qe qp ke e2 numbers of electrons 2 r2 4R2 where e is the magnitude of the charge of a single electron proton and R is the radius of the Earth 4 PSE6 23 P 009 a It is an attractive force given by the equation F k e q1 q 2 r2 b After the system comes to equilibrium the net charge is q1 q2 and they should be distributed equally in two spheres Therefore each sphere has charge q1 q2 2 Use the same expression used in part a to find the new electric force between two spheres But now the force is repulsion since both spheres have like charges 5 PSE6 23 P 018 The electric field due to the 7 00 C charge is E7 C ke 7 00 10 6 C sin 60 i cos 60 j L2 The electric field due to the 4 00 C charge is E 4 C ke 4 00 10 6 C i L2 Add up two electric fields to get the resultant E field at the position of charge q b F qE 2 6 PSE6 23 P 021 a E kqi kCq kAq kBq 2 i sin 45 i cos 45 j 2 j 2 2 ri a 2a a The magnitude of the electric field is r Ei2 Ej2 and the angle is tan Ej Ei b F qE at the same angle as in part a 7 PSE6 23 P 042 For proton it experiences a force in the same directions as E field with magnitude F qE This force accelerates the proton Therefore F qE mp ap Solve for the acceleration of the proton ap Since the proton starts out at rest vf p ap t qEt mp Similarly vf e ae t qEt me and it is in the direction opposite to the electric field because electron has negative charge 8 PSE6 23 P 054 First draw a free body diagram There are three forces present in this problem One is the gravitational force Fg pointing downward One is the tension T along the string The other one is the electric force Fe pointing to the right Since the ball is in equilibrium these three forces should balance out From Fy 0 get Fg mg T cos Solve for T From Fx 0 get Fe qE T sin Solve for q 9 PSE6 23 P 057 Label the other three point charges as follows 1 upper left corner 2 upper right corner and 3 lower right corner The forces due to these three charges are F1 F2 ke q 2 j W2 ke q 2 L W 2 i j 2 W L W2 L2 W 2 3 ke q 2 i L2 Therefore the net force is F F1 F2 F3 Use pythagorean theorem to calculate the resultant force and the angle F3 10 PSE6 23 P 062 Draw a free body diagram to locate all the forces in one of the spheres For the positively charged sphere there are four forces acting on it The first one is the gravitational force acting downward Fg mg The second one is the tension along the string T The third one is the electric force to the right Fe qE The last one is the attractive force to the left due to the negatively charged sphere 2 Fa kre2q where r 2l sin and l is the length of the string Since the sphere is at equilibrium all the forces should balance out So mg T cos qE Fa T sin Solve for E 11 PSE6 23 QQ 002 Objects A and C possess charges of the same sign 12 PSE6 23 QQ 006 It is unaffected 13 PSE6 23 QQ 007 A B C 14 PSE6 23 QQ 008 The false statement is electric field line can never intersect with one another 4
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