## College Physics (4th Edition)

$\Delta V = 150~J/C$
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: $K_2+U_2 = K_1+U_1$ $U_2-U_1 = K_1-K_2$ $(\Delta V)~q = 0-\frac{1}{2}m~v_2^2$ $\Delta V = \frac{-\frac{1}{2}m~v_2^2}{q}$ $\Delta V = \frac{-\frac{1}{2}(9.1\times 10^{-31}~kg)(7.26\times 10^6~m/s)^2}{-1.6\times 10^{-19}~C}$ $\Delta V = 150~J/C$