91Ó°ÊÓ

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 $\mathbf{5}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ }\mathbf{\text{m}}$, 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 $\mathbf{3}\mathbf{×}{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{\text{V/m}}$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.

Step by step solution

01

Identification of given data

Electric field at STP is$3×{10}^6 \text{V/m}$

02

Significance of electric field

The electric force per unit charge is referred to as the electric field. It is assumed that the field's direction corresponds to the force it would apply to a positive test charge. From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.

03

(e) Determining the ratio of the magnitude of the field you calculated in the previous part to the observed value at STP

Electric field at STP is$3×{10}^6 \text{V/m}$

And the electric field obtain from previous part is $2.87×{10}^{7 }\text{V/m}$

Ratio of the magnitude of the field you calculated in the previous part to the observed value at STP,

$\begin{array}{rcl}\text{ratio} & =& \frac{2.87×{10}^{7 }\text{V/m}}{3×{10}^6 \text{V/m}}\\ & =& 9.56\end{array}$

Hence,the ratio of the magnitude of the field youcalculated in the previous part to the observed value at STP is$9.56$

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