An electron starts at the edge of a region with a uniform electric field. The re

Question

An electron starts at the edge of a region with a uniform electric field. The region is between two parallel metallic plates. The electron has an initial velocity of 5.00 x 106m/s at an angle of 45o relative to the plane of the lower plate. The plates are 2.00 cm apart and are very large. The electric field is from the lower plate towards the upper plate and has a value of 3.50 x 103N/C. On which plate and at what location will the electron strike?

An electron starts at the edge of a region with a uniform electric field. The region is between two parallel metallic plates. The electron has an initial velocity of 5.00 x 106m/s at an angle of 45o relative to the plane of the lower plate. The plates are 2.00 cm apart and are very large. The electric field is from the lower plate towards the upper plate and has a value of 3.50 x 103N/C. On which plate and at what location will the electron strike?

Explanation / Answer

Given v0 = 5 * 10^6 m/s

theta = 45 deg

electric field E = 3.5 * 10^3 N/C

the electron experience a force in opposite direction to the field so

the acceleration of electron is in downward direction

acceleration is given by

a = F / m = q * E / m

a = 1.6 * 10^-19 * 3.5 * 10^3 / (9.1 * 10^-31)

a = 6.15 * 10^14 m/s^2 (vertical acceleration)

horizontal acceleration is 'zero'

Now this is similar to the projectile motion ( where vertical acceleration is 'g')

maximum height is given by

h = v0^2 * sin^2(theta) / 2a

h = (5 * 10^6)^2 * sin^2(45) / (2*6.15 * 10^14)

h = 1.016 cm

so this will not strike to the upper plate ( because seperation is 2cm)

the horizontal range is

R = v0^2 * sin(2theta) / a

R = (5 * 10^6)^2 * sin(2*45) / (6.15 * 10^14)

R = 4.06 cm

It strikes on the lower plate at a distance of 4.06 cm

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