My Notes 4 points I Previous Answers OSColPhys 8.8.058 7. My Notes Ask Your Teac

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My Notes 4 points I Previous Answers OSColPhys 8.8.058 7. My Notes Ask Your Teacher An electron has an initial velocity of 6.00 x 106 m/s in a uniform 4.00 x 10 Nyc strength electric field. The field accelerates the electron in the direction opposite to its initial velocity (a) What is the direction of the electric field? same direction as the electron's initial velocity (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron's speed when it returns to its starting point? m/s Additional Materials Reading

Explanation / Answer

a)

as the electric force on the electron is oposite to the electric field ,

hemce , the direction of electric field in the same direction as the electron's initial velocity.

b)

Here , acceleration of electron is given as

a = 1.602 *10^-19 * 4 *10^5/(9.1 *10^-31)

a = 7.04 *10^16 m/s^2

Now , using third equation of motion ,

d = v^2/(2a)

d = (6 *10^6)^2/(2 * 7.04 *10^16 )

d = 2.56 *10^-4 m

the distance electron travel before stoping is 2.56 *10^-4 m

c)

using first equation of motion ,

6 *10^6 = 7.04 *10^16 * t

t = 8.52 *10^-11 s

the time taken to stop is 8.52 *10^-11 s

d)

as electric field is a conservative field ,

the speed at starting point is

v = 6 *10^6 m/s

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