A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey

1.875 x 10⁶ m /s .

Explanation:

Force on electron = E e where E is electric field and e is charge on electron

acceleration generated = Ee / m where m is mass of the electron .

Putting the values

acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹

= .879 x 10¹² m /s²

v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m

v² = 0 + 2 x .879 x 10¹² x 2

v = 1.875 x 10⁶ m /s .

the answer for apex is the lowest

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