An electron is accelerated by a constant electric field of magnitude 270 N/C. (a) Find the...


An electron is accelerated by a constant electric field of magnitude 270 N/C.

(a) Find the acceleration of the electron. =..............{eq}m/s^2 {/eq}

(b) Use the equations of motion with constant acceleration to find the electron's speed after {eq}8.50 \times10^{-9}\; s {/eq}, assuming it starts from rest. =............m/s

Equation of Motion:

An equation of motion is another term used for the summation of forces or Newton's second law. If we have a sum of forces, we can use the second law to determine a time-dependent function on the position and velocity of the object.

Answer and Explanation:


  • {eq}\displaystyle E = 270\ N/C {/eq} is the electric field strength
  • {eq}\displaystyle t = 8.5\ \times\ 10^{-9}\ s {/eq} is the time elapsed

(a) Electric force can be expressed in the terms of:

{eq}\displaystyle F = qE {/eq}

Let us also recall that the force according to the second law is also written as:

{eq}\displaystyle F = ma {/eq}

Combining these two forces gives us:

{eq}\displaystyle ma = qE {/eq}

We isolate the acceleration here:

{eq}\displaystyle a = \frac{qE}{m} {/eq}

For an electron, the charge and mass is:

  • {eq}\displaystyle q = 1.602\ \times\ 10^{-19}\ C {/eq}
  • {eq}\displaystyle m = 9.109\ \times\ 10^{-31}\ kg {/eq}

So we substitute:

{eq}\displaystyle a = \frac{(1.602\ \times\ 10^{-19}\ C)(270\ N/C)}{9.109\ \times\ 10^{-31}\ kg} {/eq}

We thus get:

{eq}\displaystyle \boxed{a = 4.75\ \times\ 10^{13}\ m/s^2} {/eq}

(b) So using this acceleration, we can determine the speed by integration:

{eq}\displaystyle v = \int a dt {/eq}

We integrate this to get:

{eq}\displaystyle v = at + C {/eq}

The constant of integration here will be our velocity at time t = 0. However since the electron starts from rest, this is also zero so we have C = 0. Therefore we have:

{eq}\displaystyle v(t) = at {/eq}

We substitute:

{eq}\displaystyle v(t) = ( 4.75\ \times\ 10^{13}\ m/s^2)(8.5\ \times\ 10^{-9}\ s) {/eq}

We will get:

{eq}\displaystyle \boxed{v(t) = 403,750\ m/s} {/eq}

Learn more about this topic:

Electric Fields Practice Problems

from Physics 101: Intro to Physics

Chapter 17 / Lesson 8

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