A proton is placed in an electric field of intensity 700 N/C.What are the magnitude and direction of the acceleration of this proton due to this field? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19C) A) 6.71 × 10⁹ m/s² opposite to the electric field B) 6.71 × 10¹⁰ m/s² opposite to the electric field C) 6.71 × 10¹⁰ m/s² in the direction of the electric field D) 67.1 × 10¹⁰ m/s² opposite to the electric field ) 67.1 × 10¹⁰ m/s² in the direction of the electric field

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Quizplus · Accepted Answer

The force on the proton is F = eE = (1.60 × 10^-19 C)(700 N/C) = 1.12 × 10^-16 N. The acceleration is a = F/m = (1.12 × 10^-16 N) / (1.67 × 10^-27 kg) = 6.71 × 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