A proton is placed in an electric field E = 800 N/C.What are the magnitude and direction of the acceleration of the proton due to this field? (e = 1.60 × 10^-19C,m_proton = 1.67 × 10^-27 kg)

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The force on the proton due to the electric field is given by $F = qE$, where $q = 1.60 \times 10^{-19} \,C$ is the charge of the proton and $E = 800 \,N/C$ is the electric field. Thus, $F = (1.60 \times 10^{-19} \,C)(800 \,N/C) = 1.28 \times 10^{-16} \,N$. The acceleration $a$ of the proton is given by $a = F/m$, where $m = 1.67 \times 10^{-27} \,kg$ is the mass of the proton. Therefore, $a = (1.28 \times 10^{-16} \,N) / (1.67 \times 10^{-27} \,kg) = 7.66 \times 10^{10} \,m/s^2$. Since the proton is positively charged, it accelerates in the direction of the electric field.

A Proton Is Placed in an Electric Field E = 10910 ^ { 9 }109

A Proton Is Placed in an Electric Field E = $10 ^ { 9 }$