The electric field points due east and has a magnitude of 8.9 times 105 N/C. The

Question

The electric field points due east and has a magnitude of 8.9 times 105 N/C. The mass of the electron is 9.11 times 10-31 kg and that of the proton is 1.67 times 10-27 kg. Both the proton and electron have charge magnitudes of 1.60 times 10-19 C. What is the algebraic expression for the magnitude a of the acceleration that a charge experiences when placed in an electric field? Express your answer in terms of the magnitude |q| and mass m of the charge, and the magnitude E of the electric field. (Answer using q, m, and E.) a = Determine the magnitude ap of the acceleration of the proton. Determine the magnitude ae of the acceleration of the electron.

Explanation / Answer

a) a = |q|*E/m b) a(proton) = 8.53*10^13 m/s^2 c) a(electron) =1.56*10^17 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---