91Ó°ÊÓ

The magnitude of electric force

$$\begin{gathered} E=2.75×{10}^3\mathrm{N}/\mathrm{C} \\ \mathrm{q}=1.6×{10}^{−19}\mathrm{C} \\ \mathrm{m}=1.67×{10}^{−27}\mathrm{kg} \end{gathered}$$

The magnitude of the electric force $\left|F\right|$exerted by the proton is

$$\begin{gathered} F=\left|q\right|E \\ =\left(1.6×{10}^{−19}\right)\left(2.75×{10}^3\right) \\ =4.4×{10}^{−16}\mathrm{N} \end{gathered}$$

Hence, the magnitude of electric force felt by the proton is$4.4×{10}^{-16}N$.

Value of acceleration using force formula

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \\ =\frac{4.4×{10}^{−16}}{1.67×{10}^{−27}} \\ a=2.63×{10}^{11}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the proton acceleration is$2.63×{10}^{11}m/s^2$.

speed of (v) of proton after time $t=1.00\mathrm{μ}s$as proton starts from the rest, therefore the electric field is in one direction, and the motion will be one direction. Using constant acceleration, the velocity will be assumed by applying Newton’s Law of motion.

$$\begin{gathered} v=v_0+at \\ v=0+\left(2.63×{10}^{11}\right)\left(1.0×{10}^{−6}\right) \\ v=2.63×{10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, Proton speed from the rest is$2.63×{10}^{5}m/s$.