I am toattally lost and need help finishing and I dont know if I did part A and

Question

I am toattally lost and need help finishing and I dont know if I did part A and B right. please explain

13. In an experiment, an uniform electric eld Ex = +3:00 103 N/C is created along the x{direction. A proton
is released from rest at x = 2:00 cm.
(a) What is the change in electric potential energy of the proton when it reaches x = 4:00 cm?

I got -60V
(b) What is the velocity of the proton at x = 4:00 cm?

I got V= 1.07 *10^-5 m/s
(c) Another proton is released at x = 9:00 cm. What is its change in potential energy when it reaches
x = 1:50 cm?
(d) Is the second proton speeding up or slowing down?
(e) If the the second protons speed has changed by 50:0%, nd its velocity when it is released.
(f) Go back to the initial conditions. What is the velocity of the proton at x = 4:00 cm if there exist a
constant frictional force of 1:5 10^16 N?

Explanation / Answer

Here,

elecric field , E = 3 *10^3 N/C

xi = 2 cm

c) as the proton is moving oposite to electric field . its potential energy will increase

change in potential energy = e * E * (0.09 - 0.015 )

change in potential energy = 1.602 *10^-19 * 3 *10^3 * 0.075

change in potential energy = 3.605 *10^-17 J

d)

as the proton is moving oposite to electric field , it will slow down

e)

let the intial speed is v

final speed = 0.5 * v

Using conservation of energy

final kinetic energy + change in potential energy = initial kinetic energy

3.605 *10^-17 + 0.5 * 1.67 *10^-27 * (v/2)^2 = 0.5 * 1.67 *10^-27 * v^2

solving for v

v = 2.39 *10^5 m/s

the initial velocity of proton is 2.39 *10^5 m/s
f)

let the final velocity is v

using work energy theorum

0.5 * 1.67 *10^-27 * (v^2 - (2.39 *10^5)^2) = - 1.5 *10^-16 * (0.09 - 0.04) - 3 *19^3 * 1.602 *10^-19 * (0.09 - 0.04)

sovling for v

v = 3.86 *10^5 i

as the velocity is imaginary , it will stop in between

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