Question

We will consider the possibility that a free electron acted on by an electric field could gain enough energy to ionize an air molecule in a collision. (a) Consider an electron that starts from rest in a region where there is an electric field (due to some charged objects nearby) whose magnitude is nearly constant. If the electron travels a distance $\mathit{d}$and the magnitude of the electric field is $\mathit{E}$ ,what isthe potential difference through which the electron travels? (Pay attention to signs: Is the electron traveling with the electric field or opposite to the electric field?) (b) What is the change in potential energy of the system in this process? (c) What is the change in the kinetic energy of the electron in this process? (d) We found the mean free path of an electron in air to be about role="math" localid="1662205184726" $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ }\mathbf{\text{m}}$, and in the previous question you calculated the energy required to knock an electron out of an atom. What is the magnitude of the electric field that would be required in order for an electron to gain sufficient kinetic energy to ionize a nitrogen molecule? (e) The electric field required to cause a spark in air is observed to be about $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{\text{V/m}}$at STP. What is the ratio of the magnitude of the field you calculated in the previous part to the observed value at STP? (f) What is it reasonable to conclude about this model of how air becomes ionized? (1) Since we used accurate numbers, this is a huge discrepancy, and the model is wrong. (2) Considering the approximations we made, this is pretty good agreement, and the model may be correct.

Short answer

(a)The potential difference through which the electron travels is $\Delta V=-E·d$and the electron travelled the opposite direction of electric field

Step-by-step solution

Identification of given data

Electron travels a distance is$d$

Magnitude of the electric field is$E$

Significance of electric potential difference

A unit positive test charge must be moved from one location to another against an electric field to determine the electrical potential difference between the two points. It is also defined as product of electric field and the distance travelled by electron

$\Delta V=-E·d$ ...(i)

Where, $\Delta V$is potential difference, $E$is the electric field and $d$is the distance traveled by electron

(a) Determining the potential difference through which the electron travels

From equation (i)

$\Delta V=-E·d$

An electron's potential change is said to be in the opposite direction of the electric field when it has a negative sign. The direction of the electric field is from positive charge to negative charge, and the direction of the flow of electrons is from the region of excess negative charge to the region of positive charge.

Hence, the potential difference through which the electron travels is$\Delta V=-E·d$

and the electron travelled the opposite direction of electric field