Question

An electron moves at 2.1 × 106 m/s into a uniform electric field of magnitude 1395 N/C. The field is parallel to the electron’s velocity and acts to decelerate the electron. How far does the electron travel before it is brought to rest? The fundamental charge is 1.602 × 10−19 C and the mass of an electron is 9.109 × 10−31 kg . Answer in units of cm.

Answer 1

0.899 cm.

Step-by-step explanation:

If there is an electric field present, this means that there is an external force acting on the electron, which is equal, by definition of electric field (as force per unit charge), as follows:

F = q*E

Applying Newton´s 2nd Law, we can say the following:

F = q*E = m*a

From this equation, we can get the value of a, as follows:

a = F/m = q*E / m (1)

Now, as the electric field is uniform, this means that the acceleration that produces is constant.

This allows us to use any of the kinematic equations.

In this case, we can use this one, that doesn´t depend on time (as it is not between the givens):

vf² – vo² = 2* a* ∆x

We know that the final state of the electron is at rest, so vf=0

Replacing by the value of the acceleration a, from (1), and solving for ∆x, we have:

∆x = - vo² / 2(-e*E/m) = vo²*m /2*e*E=

∆x = (2.1)²* 10¹² (m/s)²* 9.1*10⁻³¹ kg /2* 1.6*10⁻¹⁹ C * 1,395 N/C

∆x = 0.899 cm.