An electron with an initial velocity 2

An electron with an initial velocity v_{0}=1.5 \times 10^{5} m/s enters a region of length L=1.00 cm where it is electrically accelerated as shown in the figure. It emerges with v=5.70 \times 10^{6} m/s. What is its acceleration, assumed constant?

Solution:

As the question states that the electron is in constant acceleration, we can use the following equation to solve our problem.

v^{2}=v_0^2+2a(x-x_{0})

(This is a fundamental equation that can be found in many physics books that describe the motion of a particle with constant acceleration.)

Let us now isolate a in our equation. Thus we have:

a=\frac{v^{2}-v_0^2}{2x}

Let us now substitute the values given in our problem.

a=\frac{(5.7\times 10^{6} m/s)^{2}-(1.5\times 10^{5} m/s)^2}{2(0.010 m)}

(note that we converted 1.00cm to m)

a=1.62\times 10^{15} m/s^2

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 23.