The distance in between the capacitor plates 2cm
Let t be the time, in seconds, with which proton and electron reach negative and positive charged plates respectively.
Let mp, me be the mass and qp, qe be the charge of proton and electron respectively.
ap, ae be the acceleration of proton and electron respectively, in direction of Electric field, E Let’s say Y-direction)
Let x= vertical distance traveled by proton to reach the negatively charged plate, in cm. Hence x is the distance is where we should place the electron-proton pair initially.
Hence, the distance traveled by electron 2-x) cm
The proton and electron are accelerated to the oppositely charged plates, and the expression for the respective acceleration can be written from Newton’s second law of motion.
The external electric field acting on the proton The external electric field acting on the electron E
Hence, for proton of mass mp , the expression for second law of motion can be written as,
Here the term ‘qE’ represents the external force acting on the charged particle with a charge q in an electric field of magnitude E.
Similarly the expression for electron is,
From the above equations, the accelerations can be written as,
Hence, the distance travelled by proton in a time t seconds, x, by equations of motion
Similarly for electron, the distance traveled,
Now, to find x, the distance traveled by proton, we divide eqn.1 by eqn.2, that is,
But we know, charge of proton, charge of electron,
Hence the above expression will reduce to,
Now, mass of electron, me9.1×10-31 kg
And mass of proton, mp 1.67×10-27 kg
Substitution the above values in eqn.3, we get,
By rearranging the above expression we get,
Hence the pair should be released at a distance of 1.08×10-3 cm from the negative plate.
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