Answer
The speed of the positron will be $~~1.0\times 10^7~m/s$
Work Step by Step
We can find the change in electric potential energy if the positron moves from $V=0$ to $V = 500~V$:
$\Delta U = q~\Delta V$
$\Delta U = (1.6\times 10^{-19}~C)~(500~V-0)$
$\Delta U = 8.00\times 10^{-17}~J$
In order to reach a location with an electric potential of $500~V$ the initial kinetic energy of the positron must be at least $8.00\times 10^{-17}~J$
We can find the initial kinetic energy:
$K_i = \frac{1}{2}mv_i^2$
$K_i = \frac{1}{2}(9.109\times 10^{-31}~kg)(1.0\times 10^7~m/s)^2$
$K_i = 4.55\times 10^{-17}~J$
The positron does not have enough mechanical energy to reach a location with an electric potential of $500~V$
Therefore, the positron will emerge from the field at $~~x = 0$
By conservation of energy, the kinetic energy of the positron when it emerges from the field will be equal to the initial kinetic energy.
Therefore, the speed of the positron will be $~~1.0\times 10^7~m/s$.
