Question

An electron is released from rest at a negative plate of parallel-plate capacitor. If the distance across the plate Is 10mm and the potential difference across the plate is 100v, with what velocity does the electron hit the positive plate? (M=9.1x10^-31kg, e= 1.6x10^-19 C)

Answer 1

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An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 5.0 mm and the potential difference across the plate is 5.0 V, with what velocity does the electron hit the positive plate? (me = 9.1 ? 10?31 kg, e = 1.6 ? 10?19 C)

) I am getting 1.3 * 10^6 m/sec



First we have to calculate the acceleration



F = qE

Here d denotes the between the plates

We also know that E = V/d



Therefore we get F=qV/d



Also by Newton's second law



F=ma



Therefore the given equation becomes,



a = qV/md



Also we know that the mass of the electron is 9.11*10^-31 Kg



Therefore we get a = (1.6*10^-19)*(5)/(9.11*10^-31)*(5*10^-3) = 1.8*10^14 m/sec^2 (Approx)



Therefre let V ber the final speed

Initial speed u = 0 m/sec



Therefore apply third equation of motion



V^2 - u^2 = 2*a*d



From here V^2 - 0^2 = 2*1.8810^14*5*10^-3