Solve Electric Field: Find Electron Speed

[jbot2222]

im not sure how to do this...

An electron, starting from rest, is accelerated by a uniform electric field of 8.8×104 N/C that extends over a distance of 7.0 cm. Find the speed of the electron after it leaves the region of uniform electric field.

 

[MathStudent]

There are two ways of approaching this problem, one is conservation of energy, the other is motion with constant acceleration (kinematics).. Is there any you prefer or are required to do?

Do you see why this is constant acceleration?

 

[wais]

To solve for the electron speed, we can use the equation for acceleration in an electric field:

a = qE/m

Where:
a = acceleration
q = charge of the electron
E = electric field strength
m = mass of the electron

Since we are given the electric field strength and the distance over which it acts, we can also use the equation for work done by an electric field:

W = qEd

Where:
W = work done
q = charge of the electron
E = electric field strength
d = distance over which the electric field acts

We know that the work done by the electric field is equal to the change in kinetic energy of the electron, so we can set the two equations equal to each other:

qEd = 1/2mv^2

Solving for v, we get:

v = √(2qEd/m)

Plugging in the values we have:

v = √(2(1.6x10^-19 C)(8.8x10^4 N/C)(7.0x10^-2 m)/(9.11x10^-31 kg))

v = 2.26x10^6 m/s

Therefore, the speed of the electron after it leaves the region of uniform electric field is 2.26x10^6 m/s.