## College Physics (4th Edition)

(a) The electron moved to a lower potential. (b) $\Delta V = -190~J/C$
(a) By conservation of energy, the decrease in kinetic energy is equal in magnitude to the increase in electric potential energy. Since the charge on an electron is negative, the electron must have moved to a lower potential. (b) We can use conservation of energy to find the potential difference: $K_2+U_2 = K_1+U_1$ $U_2-U_1 = K_1-K_2$ $(\Delta V)~q = \frac{1}{2}m~v_1^2-\frac{1}{2}m~v_2^2$ $\Delta V = \frac{\frac{1}{2}m~(v_1^2-v_2^2)}{q}$ $\Delta V = \frac{\frac{1}{2}(9.1\times 10^{-31}~kg)[(8.50\times 10^6~m/s)^2-(2.50\times 10^6~m/s)^2]}{-1.6\times 10^{-19}~C}$ $\Delta V = -190~J/C$