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(3) An electron is projected with horizontal velocity vo 5x10s m/s into a uniform electric field of magnitude 4000 N/C pointed upwards that is presented over a horizontal distance of 2 cm. (a) Find the magnitude and direction of the force due to the electric field. Is it justifiable to ignore the gravitational force?(b) By what vertical distance will the electron be deflected when it exits the field? (c) Find the angle with respect to horizontal as it leaves the field. This sort of configuration was used by J.J. Thompson to discover the electron with a device like the one pictured. Electrodes to generate ray Electrodes to deflect ray Displacement Anode Negative plate Voltage Cathode Evacuated tube Positive plate Cathode ray

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Answer #1

(a) As electric field is pointing upwards and the charge on the electron is negative, so the force on the electron must be towards the downward direction.

The magnitude is given by F = q*E = 1.6*10^(-19)*4000 = 6.4*10^(-16) N.

It is justifiable to ignore the gravitational force, since the mass of electron is very less (of the order 10^-31), so the gravitational force would be very small as compared to the the electric force.

(b) Let electric field be in the y direction.Since initial velocity in the y direction is zero i.e electron is travelling horizontally.

Only force in the y direction is the electric field force.

Time travel to cross the field region of 2 cm is given by, t = d/v = 2*10^(-2) / 5*10^(6) = 4 nsec.

thus displacement in y direction = Sy = (1/2)*a*t*t = (1/2)* (6.4*10^(-16))*4*4*10^(-18) / (9.1*10^(-31))

Sy = 5.6 * 10^(-3) m

(c) Since the displacement is very small the angle theta,

tan (theta) = Sy/Sx = 5.6*10(-3)/2*10^(-2) = 0.28

theta = tan (theta) = 0.28 rad.

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