Answer
0.093m/s.
Work Step by Step
Use conservation of energy. The electric field points downward, from high electric potential to low electric potential. Let the electric potential at the surface be 0, so the starting height of 2.00 m is at a potential of 300 V. We also say that the surface of the Earth is at a height of 0 for gravitational PE.
$$E_i=E_f$$
$$0+mgh+qV=\frac{1}{2}mv^2+0+0$$
Solve for the final speed.
$$v=\sqrt{2\left( gh+\frac{qV}{m}\right)}$$
Find the 2 speeds.
$$v_+=\sqrt{2\left( (9.80m/s^2)(2.00m)+\frac{(6.5\times10^{-4}C)(300V)}{0.670kg}\right)}=6.3073m/s$$
$$v_-=\sqrt{2\left( (9.80m/s^2)(2.00m)+\frac{(-6.5\times10^{-4}C)(300V)}{0.670kg}\right)}=6.2143m/s$$
Calculate the difference.
$$6.3073m/s-6.2143m/s=0.093m/s$$
