## College Physics (4th Edition)

The magnitude of the acceleration is $3.16\times 10^{12}~m/s^2$ and the direction of the acceleration is straight up.
The electric field is directed straight up. Since the charge on the proton is positive, the force exerted on the proton will be directed in the same direction as the electric field. Therefore, the direction of the force exerted on the proton, and thus the direction of the acceleration, is straight up. We can find the force exerted on the proton: $F = E~q$ $F = (33,000~N/C)(1.6\times 10^{-19}~C)$ $F = 5.28\times 10^{-15}~N$ We can find the proton's acceleration: $F = ma$ $a = \frac{F}{m}$ $a = \frac{5.28\times 10^{-15}~N}{1.67\times 10^{-27}~kg}$ $a = 3.16\times 10^{12}~m/s^2$ The magnitude of the acceleration is $3.16\times 10^{12}~m/s^2$ and the direction of the acceleration is straight up.