In the figure below, a uniform, upward electric field E of magnitude 2.00 × 103
ID: 1573460 • Letter: I
Question
In the figure below, a uniform, upward electric field E of magnitude 2.00 × 103 N/C has been set up between two horizontal plates by charging the lower plate positively and he upper plate negatively. The plates have length L-8.9 cm and separation d 2.40 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity Vo of the electron makes an angle 0 32.0° with the lower plate and has a magnitude of 5.70 x 106 m/s. Which, if any plate will the electron hit? select- How far horizontally from the left edge will the electron strike? cm (Enter 0 if the particle does not hit a plate.)Explanation / Answer
a)
a = Eq/m = 2000*1.6*10^-19/(9.1*10^-31)
= 3.52*10^14 m/s2
Maximum height reached, H = u^2/2a
= (5.7*10^6*sin(32 deg))^2/(2*(3.52*10^14))
= 0.013 m = 1.3 cm
As the maximum height reached is less than ( d = 2.4 cm ), so it wont hit the upper plate.
Using the equation of motion:
s = ut + 0.5*at^2
0 = 5.7*10^6*sin(32 deg)*t + 0.5*3.52*10^14*t^2
So, t = 1.72*10^-8 s <------- time required to come back to the same level.
So, horizontal distance covered in this time = 1.72*10^-8*5.7*10^6*cos(32 deg)
= 0.083 m = 8.3 cm
As the horontal range (R = 8.3 cm) is less than L = 8.9 cm, so it will hit the lower plate <-------- answer
b)
distance from the left edge = 8.3 cm <------ answer(solved in previous part)