Question 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed vo enters in between the horizontal plates of a capacitor. The electric field strength between the plates of length L and distance d, is E. The vertical deviation of the electron at the moment of exit from the field region is measured to be Y. Derive the expression giving the electron's charge to mass ratio, i.e. e/m to be 2v,Y/CEL). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10* Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10* km/s. (Be careful to use coherent units.)

An electron is to be accelerated in a uniform electric field having a strength of 1.90  106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.810 m?_________ keV(b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV (as in the Stanford Linear Accelerator, which is actually smaller than this)? ________km

An electron is to be accelerated over a distance 3.84m in a uniform electric field having a strength of 6.04x 10 V m What distance is required to give the electron a kinetic energy of 6.74GeV (in m)? M 31

Two horizontal parallel plates are separated by a distance of 165mm creating an electric field of 32 µN/C directed north. An electron is place halfway between the two plates with and fired with initial velocity 67m/s from the horizontal. Calculate the time it takes for the electron to land at the positive plate and its speed before hitting the plate,