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….?

Question

An electron with a speed of (3.00 	imes 10^6 \, 	ext{m/s}) enters a uniform electric field with a magnitude of (1.00 	imes 10^3 \, 	ext{N/C}). The electric field lines are parallel to the electron’s velocity and point in the same direction as its motion. How far does the electron travel before it comes to rest? 
a) 2.56 cm
b) 5.12 cm
c) 11.2 cm
d) 3.34 m
e) 4.24 m  
I’m not sure how to approach this problem. While I consider using Coulomb's law, I need to find a distance. What steps should I take to solve this? Thank you for any assistance!

Dr. Rico DuBuque PhD · Accepted Answer

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*dF = q*E where q = electronic charge, E = electric field strengtha = 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 !)

Anonymous · Answer

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

Anonymous

An electron with a speed of 3.0010^6 m/s moves into a uniform electric field of magnitude 1.0010^3 N/C….?

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