Question

Cool down question number 3. Please make sure to state what direction the electron will move, and the kinetic energy after it moves 1 cm in that direction. Please show all detailed work, step by step.

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Cool Down(s) 1. Think about your work during the recent electric field lab and then consider the plot shown in figure Q21.6 on page 694 of your text. What are the magnitude and direction (to the right or to the lef of the electric field at the position x2.5mP 3. Consider figure 21.23 on page 683 in your text. Imagine an electron being released from rest at the point labeled 2. In what direction will the electron move? What is its kinetic energy after it moves 1 cm in that direction'?

Answer 1

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Electric potential decreases in the direction of electric field. so in the given figure, electric field is in left (-ve x - axis) direction. Since electron is having a negative charge on it and -ve charge moves opposite to the direction of electric field. so it will move in right direction (+ve x - axis) After moving to the right. It will be at a potential of 75 v [100 + 50/2 = 75 v] Now K. E = change in potential energy = q v_i - q v_f = -1.6 times 10^-19 times (50 - 75) = 25 times 1.6 times 10^-19 q = 1.6 times 10^-19 c v_i = 50 v v_f = 75 v rightdoublearrow K. E = 25 ev In terms of Joule, it is K. E = 4 times 10^-18 J