An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.5 cm, and the electric field within the capacitor has a magnitude of 3.9 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Explanation:

When a q charge travels in an electric field , it acquires energy which is equal to V x q where V is potential difference and q is amount of charge

Here electric field E  = V / d where V  is potential difference between plate and d is plate separation

V = E x d

= 3.9 x 10⁶ x 1.5 x 10⁻²

= 5.85 x 10⁴ V

energy of electron = V x q  

= 5.85 x 10⁴ x 1.6 x 10⁻¹⁹ J

= 9.36 x 10⁻¹⁵ J  

Kinetic energy of electron = 9.36 x 10⁻¹⁵ J

2.6 and i can kick good lol