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# Electrostatics and acceleration of an electron

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a) The direction of electric field vector E is from the +ve plate towards the -ve plate (because a unit +ve charge placed any where between the plates will travel in that direction). However, electron being negatively charged will travel from the -ve plate towards the +ve plate.

Force experienced by a charge q placed in a field E is equal to qE in magnitude. In case of the electron: F = eE

Acceleration a = Force/Mass = eE/m where m = mass of electrom = 9.11 x 10^-31 kg

Substituting values: a = (1.6 x 10^-19 x 1.0 x 10^4)/9.11 x 10^-31 = 17.6 x 10^14 m/s^2

b) Initial speed of the electron = u = 0

Final speed = v

Acceleration a = 17.6 x 10^14 m/s^2

Distance traveled = s = 1 cm (0.01 m)

Applying the equation v^2 - u^2 = 2as we get: v^2 = 2 x 17.6 x 10^14 x 0.01 = 35.2 x 10^12

v = 5.9 x 10^6 m/s

KE = ½ mv^2 = ½ x 9.11 x 10^-31 x 35.2 x 10^12 = 16 x 10^-18 J
c) Applying the equation v = u + at we get: 5.9 x 10^6 = 0 + 17.6 x 10^14 t

Time t = 3.35 x 10^-9 sec

Solution:

Magnitude of force due to each charge q_1 and q_2 on q_3 = F = kq_1 q_3/d^2 where k ...

#### Solution Summary

This solution provides detailed, step-by-step explanations to the given physics problems regarding electrostatics an acceleration of an electron.

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