An electron (mass m = 9.11E-31 kg) is accelerated in the uniform field E (E = 1.

Question

An electron (mass m = 9.11E-31 kg) is accelerated in the uniform field E (E = 1.25E+4 N/C) between two parallel charged plates. The separation of the plates is 1.40 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure below.

Link to figure:
http://spock.physast.uga.edu/res/prenhall/giancoli/Physics-Principles_and_Applications_6e/Chap16/graphics/gian1644.gif

With what speed does it leave the hole?


Show that the gravitational force can be ignored by calculating the ratio of the gravitational to the electric force. Calculate that ratio.

Explanation / Answer

Mass of the electrom m = 9.11 *10^-31 kg   Electric field E = 1.25 *10^4 N/C , distance between the plates d = 1.40cm           F = q E              = 1.6 *10^-19 C * 1.25 *10^4 N/C              = 2 .0*10^-15 N     acceleration of the electron a = F / m                                                  = 2 .0*10^-15 N / 9.11*10^-31 kg                                                  = 22.2 *10^14 m/s^2            distance between the plates is d = 1.40 cm                       V   = E *d                            = 1.25 *10^4 N/C * 1.40 *10^-2m                              = 1.75 V           1/2 mv^2 = q * V           1/ 2 * 9.11*10^-31 kg * v^2 = 1.6 *10^-19C * 1.75 V         speed of the electron is v = 0.78*10^6 m/s Gravitational force F_G = G m_e m_p / r^2      Electro static force F_e = k q_e q_p / r^2                       F_g / F_e =   G m_e m_p / k q_e q_p       substitute the values to get solution

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.