Question

In the figure a uniform, upward electric field of magnitude 1.90 × 10³ N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 13.0 cm and separation d = 1.60 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity of the electron makes an angle θ = 50.0° with the lower plate and has a magnitude of 6.40 × 10⁶ m/s. (a)Will the electron strike one of the plates? (b) If so, which plate, top or bottom? (c) How far horizontally from the left edge will the electron strike?

I know it strikes the top plate, but I can't find the answer to c. After calculating, I found a distance of 0.015 m. Which is apparently wrong.

Answer 1

c)

consider the motion in vertical direction

q = charge on electron = 1.6 x 10^-19 C

m = mass of electron = 9.1 x 10^-31 kg

E = electric field between the plates = 1900 N/C

a = acceleration of the electron

acceleration of the electron is given as

a = qE/m

a = (1.6 x 10^-19)(1900)/(9.1 x 10^-31)

a = 3.34 x 10¹⁴ m/s²

v_oy = initial velocity along the vertical direction = 6.4 x 10⁶ Sin50 = 4.9 x 10⁶ m/s

t = time taken

y = vertical displacement = 1.60 cm = 0.0160 m

using the equation

y = v_oy t + (0.5) a t²

0.0160 = (4.9 x 10⁶) t + (0.5) (- 3.34 x 10¹⁴) t²

t = 3.7 x 10^-9 sec

consider the motion along the x-direction :

v_ox = velocity along the horizontal direction = 6.4 x 10⁶ Cos50 = 4.1 x 10⁶ m/s

X = horizontal displacement

using the equation

X = v_ox t

X = (4.1 x 10⁶) (3.7 x 10^-9)

X = 0.0152 m

X = 1.52 cm