What is the strength of the electric field between two plate

What is the strength of the electric field between two plates separated by a distance 6.0 cm and with a potential difference of 2.30×10⁴ V.

Correct, computer gets: 3.83E+05 V/m

In the previous problem, if a charge of 0.10 C is released from rest at the positive plate, how much kinetic energy will it have when it reaches the negative plate?

HINT: The increase in kinetic energy must be equal to the decrease in potential energy

THE ANSWER IS NOT .0023 J or -.0023 J

Solution

Potential difference V = 2.3 x10 ⁴ volt

Distance between two plates d = 6 cm = 0.06 m

The strength of the electric field between two plates E = V/d

                                                                              =(2.3x10 ⁴) / 0.06

                                                                              = 3.83 x10 ⁵ volt/m

Charge q = 0.1 x10 ^-6 C

increase in kinetic energy = Potential energy decrease

(1/2) m[v ² -u ²] = Vq          Where u = initial velocity = 0

                       =[2.3x10 ⁴x0.1x10 ^-6 ]

                       = 2.3x10 ^-3 J

                       = 2.3 mJ