A uniform, upward electric field E of magnitude 2.00 x 10^3 N/C has been set up

Question

A uniform, upward electric field E of magnitude 2.00 x 10^3 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 = 10.0 cm and separation d = 2 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle theta = 45 with the lower plate and has a magnitude of 6 x 10^6 m/s. a) Will the electron strike one of the plates? b) If so, which plate and how far horizontally from the left edge will the electron strike?

Please explain all the steps.

Explanation / Answer

Force on elctron is = q E which is e E where e is the electron charge.

Acceleration = e E / m so it is in the downward direction.

a = 1.75 * 10^11 * 2 * 10^3 = 3.5 * 10^14 m/s^2

This is just like a projectile motion.

h maximum = v^2 sin^2 (theta) /2 a

So maximum h = 36 *10^12 * (0.5) / 2 * 3.5 * 10^14 = 0.0257 m = 2.57 cm

So the electron strikes upper plate.

0.02 = 6 * 10^6 * sqrt(0.5) *t - 0.5 * 3.5 * 10^14 *t^2

t = 6.4 * 10^-9 seconds

So distance traveled = 6 * 10^6 * sqrt (0.5) * 6.4*10^-9 = 2.715 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.