Answer
$\Delta V = 150~J/C$
Work Step by Step
By conservation of energy, the increase in kinetic energy is equal in magnitude to the decrease in electric potential energy. We can use conservation of energy to find the potential difference:
$K_2+U_2 = K_1+U_1$
$U_2-U_1 = K_1-K_2$
$(\Delta V)~q = 0-\frac{1}{2}m~v_2^2$
$\Delta V = \frac{-\frac{1}{2}m~v_2^2}{q}$
$\Delta V = \frac{-\frac{1}{2}(9.1\times 10^{-31}~kg)(7.26\times 10^6~m/s)^2}{-1.6\times 10^{-19}~C}$
$\Delta V = 150~J/C$
