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Question 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed v0 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 2v0^2Y/(EL^2). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10^4 Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10^4 km/s. (Be careful to use coherent units.)

          Question 1
a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed v0 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 2v0^2Y/(EL^2). (Recall that Thomson received Nobel Prize for his achievement.)
b) Calculate e/m, knowing the following data. E=1.6x10^4 Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10^4 km/s. (Be careful to use coherent units.)

Question 1
a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed v0 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 2v0^2Y/(EL^2). (Recall that Thomson received Nobel Prize for his achievement.)
b) Calculate e/m, knowing the following data. E=1.6x10^4 Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10^4 km/s. (Be careful to use coherent units.)

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