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.

0.899 cm.

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.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

continuous conducting path for current, power supply