A proton is placed in an electric field of intensity $800 \mathrm {~N} / \mathrm { C }$. What are the magnitude and direction of the acceleration of the proton due to this field? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , \mathrm { m } _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)$

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The force on the proton is given by $F = Eq$, where $E = 800 \, \mathrm{N/C}$ and $q = 1.60 \times 10^{-19} \, \mathrm{C}$. The acceleration $a$ is found using $F = ma$, leading to $a = \frac{F}{m} = \frac{Eq}{m}$. Substituting the given values, $a = \frac{(800)(1.60 \times 10^{-19})}{1.67 \times 10^{-27}} = 7.66 \times 10^{10} \, \mathrm{m/s^2}$. Since the proton is positively charged, it accelerates in the direction of the electric field.

A Proton Is Placed in an Electric Field of Intensity 800 N/C800 \mathrm {~N} / \mathrm { C }800 N/C

A Proton Is Placed in an Electric Field of Intensity $800 \mathrm {~N} / \mathrm { C }$