Question

Determine the magnitude of the electrical force Fo one that the object exerts on the electron. Express your answer with the appropriate units. ReviewI Constants | Periodic Table An electron is 0.18 cm from an object with electric charge of +3.0 x 10-6 C 1.33.10-6 N Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining Part B The electron is pulled so that it moves to a distance of 0.19 cm from the charged object. Determine the magnitude of the electrical force Fo on e exerted by the charged object on the electron when at this distance. Express your answer with the appropriate units. Units Submit

Answer 1

Part A Write the equation of Coulomb's Law, F = 59,92 Here, q, and q, are charges and is the distance between them. k is coulomb's constant. Substitute -1.6x10-19 C for 91, +3x10-C for 92, 0.18x102 m forr and 9x109 Nm²/C2 for Focne =(9x10° Nm²/C) (-1.6x10-6C)(+3x10-c) (0.18x102 m) =-1.33x10-9N Hence, the magnitude of the force on the electron is Fome = 1.33x10-N. Part B Again, write the equation for Coulomb's Law, F = k 9192 Substitute 0.19x10-2 m forr, F'octe = (9x10° Nm²/C2)(-1.6x10° C)(+3x10^c) (0.19x102 m) =-1.197 X10-9 Hence, the magnitude of the electric force is Foone = 1.20x10-°N

Part C Work done by the average force pulling the electron is Fone+F One Ar 2 Here, Ar =(0.19 cm -0.18 cm) =10*m Substitute the value 1.33x10- N for Foone, 1.20x10-° C for Foone and 10+ m for Ar in work equation. (1.33x10 N+1.20x10°N =1.265 x 10-13 J Hence, the work done by the average force is W = 1.265 x10- J.

Part D Electric potential energy at the initial position, U; = 9192 Electric potential at the final position, U2 = k 4.92 Thus the change in potential energy, AU = U2-U =kg.g. ( In Substitute 0.19x10 m forr, and 0.18x102 m for , AU, = (9x109 Nm²/C°)(-1.6x10–)(+3x10^ c) 0,19x102m- = 1.263x10-13 J L0.19x10 m 0.18x10 m Hence, the change in electric potential energy is AU, = 1.263x10-13 J.