A proton is placed in an electric field of intensity 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) A)7.66 × $10 ^ { 9 }$ m/ $\mathrm { s } ^ { 2 }$ opposite to the electric field B)7.66 × $10 ^ { 10 }$ m/ $\mathrm { s } ^ { 2 }$ opposite to the electric field C)7.66 × $10 ^ { 10 }$ m/ $\mathrm { s } ^ { 2 }$ in the direction of the electric field D)76.6 × $10 ^ { 10 }$ m/ $\mathrm { s } ^ { 2 }$ opposite to the electric field E)76.6 × $10 ^ { 10 }$ m/ $\mathrm { s } ^ { 2 }$ in the direction of the electric field

Question

Quizplus · Accepted Answer

The force on the proton is given by F = eE = (1.60 × 10^-19 C)(800 N/C) = 1.28 × 10^-16 N. The acceleration is a = F/m = (1.28 × 10^-16 N) / (1.67 × 10^-27 kg) = 7.66 × 10^10 m/s². 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 10910 ^ { 9 }109

A Proton Is Placed in an Electric Field of Intensity $10 ^ { 9 }$