In the figure below, a uniform, upward electric field B of magnitude 2.00 x 10 N

Question

In the figure below, a uniform, upward electric field B of magnitude 2.00 x 10 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 = 1.85 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 -51.00 with the lower plate and has a magnitude of 6.10 × 106 m/s. I. Which, if any plate will the electron hit? How far horizontally from the left edge will the electron strike? cm (Enter 0 if the particle does not hit a plate.)

Explanation / Answer

The y component of the velocity = 6.1e6 sin(51) = 4.74 x 10^6 m/sec

Acceleration in y direction = Force / mass = qE/m = 2e3 x 1.6e-19 / 9.1e-31 = 3.5e14 m/sec

So time taken to cover vertical distance = .0185 = 4.74 x 10^6 x t + .5 x 3.5e14 x t^2

So, t = 3.45e-9 sec

The horizontal distance coveredd = 3.45e-9 x 6.1e6 cos(51) = .0132 m = 13.2 cm

So it will miss the plate.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---