# An electron with a speed of 3.00*10^6 m/s moves into a uniform electric field of magnitude 1.00*10^3 N/C….?

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An electron with a speed of 3.00*10^6 m/s moves into a uniform electric field of magnitude 1.00*10^3 N/C. The field lines are parallel to the electron’s velocity and point in the same direction as the velocity. How far does the electron travel before it is brought to rest? a)2.56cm b)5.12cm c)11.2cm d)3.34m e)4.24m

I’m not really sure what to do here. I could use coulomb’s law but I need to find a distance…how would I do that? Thanks for any help. : )

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• To answer this question you find the electrostatic force (F) on the electron, then find its acceleration (a), and then use the formula for uniformly accelerated motion v^2 – u^2 = 2*a*d

F = q*E where q = electronic charge, E = electric field strength

a = F/m = q*E/m where m = mass of electron.

d = u^2/(2*a) = m*u^2/(2*q*E) where u is the original speed of the particle.

I leave the arithmetic to you !…. (I make it 2.56cm, but you had better check !)

• the electrical powered field vector is defined because of the fact the rigidity that could desire to be exerted on a unit cost ought to the cost be placed at that element. you are able to hence say F=|E|q |E|q=ma a=|E|q/m The proton will advance as much as the left, collectively as the electron will advance as much as the perfect because of the fact the electrical powered field factors in the path that a advantageous cost will advance up in direction of. in basic terms plug in for acceleration and use v_f^2-v_i^2=2a*distance to locate v_f. sturdy success! =D 