an electron accelerates from rest in a uniformelectric field to a speed of 2900000m/s(2.9*10^6m/s) in a distan

Question

an electron accelerates from rest in a uniformelectric field to a speed of 2900000m/s(2.9*10^6m/s) in a distance of 14cm. What is the magnitude of the electric field?

Answer 1

find acceleration a, using that final velocity,v, and the distance,s, to accelerate,a, the electron to that velocity (use v²=2as). then use the mass of electron 9.11E-31 kg to find the force required to accelerate it using F=ma. F is yielded in newtons.

according to coulombs law, F=qL, where q is the charge of electron, 1.60E-19 Coulombs, L is the electric field in N/C or V/m units, whichever you prefer.

Answer 2

get the electron rest mass from somewhere. Claim that the final speed is << c so you can ignore relativistic effects - calculate the kinetic energy in the electron at the final speed. This energy was achieved by doing work, applying a constant force over a 0.014m distance. Find out what the force was.

Best of Luck - Mike

Answer 3

F=ma
F=qE

so, qE = ma
or, E = ma/q
again v^2 = u^2 + 2as = 2as
or, a = v^2 / 2s
or, a = (2.9*10^6)^2 / 2/ 0.14
or, a = 3003.57*10^10

So, E = 9.1*10^(-31) * 3003.57*10^10 / 1.6E-19

or, E = 5.88*10^-5