Calculating Time for Electron to Slow Down in Electric Field

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An electron with an initial velocity of 2.4 × 10^6 m/s in an electric field of 1250 N/C will slow down due to its negative charge. The force experienced by the electron is calculated using F = E * q, resulting in a force of -2.0025 × 10^-16 N. This leads to an acceleration of -2.2 × 10^14 m/s² when applying Newton's second law. The time required for the electron's speed to reduce to one fourth of its original value is determined to be approximately 3 × 10^-9 seconds, although some calculations suggest it could also be 8 × 10^-9 seconds. The discussion emphasizes using differential equations to relate acceleration and velocity over time.
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Homework Statement



An electron with velocity v = 2.4 × 10^6 m/s i encounters an electric field E = 1250 N/C i. How long will it take for the speed to be one fourth its original value?

Homework Equations



Not sure

Maybe E = kQ/r^2

The Attempt at a Solution



I don't know where to begin. Looking at it makes me think it won't slow down because both are pointing in the positive i direction. Also, does an electron have a charge of -1.602 X 10 ^-19 C?
 
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The electron will slow down. Given some electric field and a charge, what is the force that will be experienced by the electron? Knowing this, how can you calculate the acceleration using Newton's 2nd Law? This should give you a good start. Your equation for an electric field is not needed since you are told what the (constant) electric field is for this problem.
 
Also, does an electron have a charge of -1.602 X 10 ^-19 C?
Right. And as it has a negative charge, it will slow down when it moves in the direction of the field lines.
Can you relate the electric field to a force on the electron?

Maybe E = kQ/r^2
This does not help.
 
So what I did was multiply E and the charge of an electron to find the force between them.
F= -2.0025 X 10^-16 N
Then using F=ma
when m = 9.109 X 10^-31 kg
I got a = -2.2 X 10^14 m/s^2

First, is this the direction I need to be going in or am I doing it wrong?
Second, if this is right, how can I use the a to find the velocity and time it takes?

Can I do the differential equation
dv/dt=a?
 
Last edited:
That is a good approach.
Can I do the differential equation
dv/dt=a?
You can use this to find the velocity as function of time, right. As the acceleration is constant, this is easy to solve.
 
I got the time for it to get to a fourth of the velocity to be 3 X 10^-9 s

That sound right? Is there a constant from the integration or will it be zero?

I also think it might be 8 X 10^-9 s by doing
v(0)=a(0)+c
v(0)= initial velocity
So c= the initial velocity.
 

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