Question

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due to the air in the cloud and any collisions with air molecules. (a) What acceleration does the maximum electric field produce for protons in the cloud? Express your answer in SI units and as a fraction of g. (b) If the electric field remains constant, how far will the proton have to travel to reach 10% of the speed of light ( 3.00 × 10 8 m/s) if it started with negligible speed? (c) Can you neglect the effects of gravity? Explain your answer.

Answer 1

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.