Magnitude of An Electric Field?

Question

Suppose a proton accelerator is 4.0 m long and must accelerate protons from rest to a speed of 1.0 x 10^7 m/s.

What is the magnitude of the electric field that could accelerate these protons?

Answer 1

Let's put some equations down
v = at
d = 1/2 at^2
let's put in what we know
10^7 = at
4 = 1/2 at^2

a=10^7/t
4=1/2.t^2.10^7/t -> t=8/10^7
a = 10^14/8 = 1.25 x 10^13 m/s^2
t = 8 x 10^-7 s

So from this acceleration we can deduce
using F = ma, knowing that the mass of a proton is 1,6726 x 10^(-27) kg
so F = 1,6726 x 10^(-27)kg * 1.25 x 10^13 m/s^2 = 2.09075 x 10^-14 N

This is the force required. The proton has a charge of 1.602 × 10^−19 C

And the voltage required is

V = F.d/q = 2.09075E-14*4/1.602E−19 =
523720V = 524 kV

Answer 2

Kinetic energy gain of proton = 1/2 m v^2. This comes from work done on charge = qV. If E =V/d then V =Ed
So qEd = 1/2 m v^2 then
E =1/2 (1.67 x 10^-27) 10^14 /[(1.6 x 10^-19)x4]

= 130 kV/m

Answer 3

F = q E
we want E. let's ignore relativistic effects. you have a proton that needs to go from 0 to 1E7 m/s in 4 m. so,
4m = 0.5 a t^2
1E7 m/s = a t

8 = a t^2
1E7 / t = a

8 = (1E7 / t) (t^2)
8 = 1E7 t
t=8E-7s
a = 1E7/8E-7 = 1.25E+13 m/s^2
F = m * a = (mass of proton) * 1.25E+13 m/s^2
F = 2.09078E-14 N = q E = 1.602E-19 C * E
E = 130510.4229 newtons per coulomb

Answer 4

electric powered container is 0 i think of. to paintings out the sphere, take each and every can charge in my view; E = (a million/4*pi*e)*Q*root2/2 And superimpose. i won't bear in mind a thank you to calculate electric powered ability on the 2nd. i understand that container = gradient of ability.