In the figure a uniform, upward electric field of magnitude 1.80 × 10 3 N/C has

Question

In the figure a uniform, upward electric field of magnitude 1.80 × 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 = 11.0 cm and separation d = 1.70 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 ? = 47.0° with the lower plate and has a magnitude of 6.70 × 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?

Explanation / Answer

To determine whether or not the electron strikes one of the plates, we need to determine the time Ty required to travel a vertical distance of y = 0.017 m and the time Tx for a horizontal distance of x = 0.11 m.

If Ty < Tx, then the electron will strike the negative plate.
If Ty > Tx, the electron will not strike the plate and we will then determine the vertical distance at which the particle leaves the space between the plates.

For the most part, this is a kinematics problem, but we need to evaluate the vertical acceleration induced on the electron as it travels through the plates.

This acceleration is found by equating F = qE = ma --> a = qE/m = (1.6e-19)(1.8e3)/(9.11e-31) = 3.16e14 m/s^2.

We also need to isolate the x and y components of the velocity v0.

Vy = v0sin(47) = 4.9e6 m/s
Vx = v0cos(47) = 4.56e6 m/s

Now we find Ty and Tx.

0.017 = 0+(4.9e6)(Ty)+(0.5)( 3.16e14)(Ty)^2 --> Ty = 3.15e-9 s
0.11 = 0+(4.56e6)(Tx) --> Tx = 2.4e-8 s

Since Ty < Tx, the electron will in fact strike the top plate at a horizontal distance of x = 0+(4.56e6)(3.15e-9) = 0.014 m.

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