Electrons are accelerated to reach a kinetic energy of 250 eV (= 4.0 x 10-17 J). If the acceleration is done in a distance of 0.20 m, what is the magnitude of the electric field causing the acceleration?

Answer 1

KE= m•v²/2 => v²= 2•(KE)/m
a=v²/2•s =2(KE)/2•m•s = KE/m•s
m•a=e•E
E= m•a/e = m•KE/e•m•s = KE/e•s =
=250•1.6•10⁻¹⁹/1.6•10⁻¹⁹•0.2 = 1250 V/m