An electron, initially at rest is placed in an electric field of 5x10 N/C directed west. What is the acceleration of the electron due to the electric field? O 4.79x10^13 ms^2 O 8.78x10^-16 ms^2 8.78x10^16 m's^2 O 4.79x10^-13 ms^ 2

An electron moves initially to the right as it enters a region between two parallel plates that have charges of equal magnitude and opposite sign. The initial horizontally directed speed is 6.00 x 106 m/s as shown by the arrow at the left end of the plates. The uniform electric field between the plates has magnitude E 12.00 N/C. The time the electron spends between the plates is t 0.500 µs. When the electron finally exits the region between the plates, the electron has picked up a y-component of velocity Vy shown by the vertical arrow at the right end of the plates. Let the positive y-direction be upward and the positive x-direction be rightward. (i) Which plate is positive, the bottom or the top one? Explain. (ii) What is the direction of the electric field between up or down? Explain. (iii) What is the value of Vy? The electron mass is m= 9.11x10-31 kg and the charge is -1.60x10-19 C. (iv) What is the length L of the plates?

Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 305 N/C. (a) What is the magnitude of the acceleration of the electron (in m/s2)?                                        m/s2 (b) The electron is initially at rest. What is its speed (in m/s) after 1.20 ✕ 10−8 s?                                        m/s

An electron with speed v0 = 3388829.6m/s travels parallel to and along the same direction as an electric field of magnitude E = 3154N/C. (a) How far will the electron travel before it stops? Answer in SI units and multiply your answer by 10^2. (b) How much time will elapse before it returns to its starting point? Answer in SI units and multiply your answer by 10^8.