Question

The practical limit to an electric field in air is about $3.00 \times 10^6 \textrm{ N/C}$. Above this strength, sparking takes place because air begins to ionize and charges flow, reducing the field. (a) Calculate the distance a free proton must travel in this field to reach 3.00% of the speed of light, starting from rest. (b) Is this practical in air, or must it occur in a vacuum?

- $0.141 \textrm{ m}$
- No, this is not practical in air since the proton would collide with many air particles while traveling this distance.

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This is College Physics Answers with Shaun Dychko. We're going to calculate the distance a proton needs to travel to reach a speed which is three percent the speed of light, which is 0.03 times the speed of light which is abbreviated with the letter c all the time. So that's 0.03 times three times ten to the eight meter per second. And that is nine times ten to the eight meters per second. How far will it have to travel to get that speed in the presence of an electric field of magnitude three times ten to the six newtons per Coulomb assuming it starts from rest. So we have this Kinematic Formula that the final velocity equals initial velocity plus two times acceleration times the distance. And we'll solve for

*d*, because we know the initial speed is zero and so we'll divide both sides by two*a*and get*d*as the final speed squared over two*a*. Now, we'll figure out acceleration based on newtons second law that net force is mass times acceleration and we'll divide both sides by*m*. And solve for*a*is force divided by mass. And the force in this case is that provided by the electric field and that is electric field strength times the charge. So we substitute*Eq*in place of an*F*. And then we substitute all of this in four*a*. And since we're dividing by*a*, it's a bit messy to divide a fraction in a fraction, so I'm going to multiply by the reciprocal of*a*which is the same as dividing by*a*. So multiplying by*m*over*Eq*. And so the distance is going to be, the final speed nine times ten to the six meters per second squared times the mass of a proton divided by two times this electric field strength that we're given three times ten to the six newtons per Coulomb times by the elementary charge on the proton. And we get 0.141 meters. And in part b, this is not a practical thing to happen in air since the proton would actually collide with many air particles while traveling this distance and this formula assumes constant acceleration but when the proton collides with an air particle, it will be accelerated in lots of different directions and so the acceleration will not be constant.