Answer
0.25 N/C
Work Step by Step
Assume that this is the only force acting on the proton. Apply Newton’s second law, and use equation 16–3 to calculate the magnitude of the field.
$$\vec{E}=\frac{\vec{F}}{q}$$
$$\vec{F}=q\vec{E}=m\vec{a}$$
$$E=\frac{ma}{q}=\frac{(1.673\times10^{-27}kg)(2.4\times 10^6)(9.80\;m/s^2)}{1.602\times10^{-19}C}$$
$$= 0.25N/C$$
