Answer:
E= 50.1*10³ N/C
Explanation:
Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:
[tex]vf^{2} -vo^{2} = 2*a*x[/tex]
We know that v₀ = 0 (it starts from rest), that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:
[tex]a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2[/tex]
According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.
Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:
F = q*E (1)
In this case, q=e= 1.6*10⁻19 C
But this force, can be expressed in this way, according Newton's 2nd Law:
F = m*a (2) ,
where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.
From (1) and (2), we can solve for E, as follows:
[tex]E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C[/tex]
⇒ E = 50.1*10³ N/C
