A uniform, upward-pointing electric field E of magnitude4.00×10 3 N/C has been s

Question

A uniform, upward-pointing electric field E of magnitude4.00×103N/C has been set up between twohorizontal plates by charging the lower plate positively and theupper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. Electrons are shot between the platesfrom the left edge of the lower plate.
The first electron has the initial velocityv0, which makes an angle=45° with the lower plate and has a magnitude of5.52×106m/s. Will this electronstrike one of the plates? If so, what is the horizontal distancefrom the left edge? If not enter the vertical position at which theparticle leaves the space between the plates.

Help Please! Have been working for hours but can't get the rightanswer!

Explanation / Answer

Look at some of the charateristics of the problem v0y = v0x = v0 cos45     to get the time for the electron tocross the plates. v0y = a t      is the timefor the particle to reach its maximum possible height. (and thetime to get back to the lower plate) This will give you an idea whether the particle reaches theupper plate or not and where it is relative to its horizontalposition.. a = F / m = E q / m   is the acceleration (downwardof the electron) Now with this information you should be able to applykinematic relations to solve the problem. For instance: sy = v0y t - 1/2 at2    Also, you have the range formula R = v02/ g       (sin 2 =1    and g = a here) v0y = v0x = v0 cos 45 =3.90 * 10E6 m/s a = F / m = E q / m = 4 * 10E3 * 1.6 * 10E-19 / 9.11 * 10E-31= 7.025 * 10E14 m/s2 th = 3.9 * 10E6 / 7.025 * 10E14 = 5.55 * 10E-9sec     time for electron to reach maximumheight tx = .04 / 3.90 * 10E6 = 1.026 * 10E-8sec        greater than time toreach max height hmax = 3.90 10E6 * 5.55 * 10E-9 - 1/2 * 7.025* 10E14 * (5.55 * 10E-9)2 = .011 m = 1.1 cm Thus the electron does not reach the plate The horizontal distance it travels is d = v0x t = 3.9 * 10E6 * 2 * 5.55 * 10E-9 =4.3 cm     the electron does not get back tothe lower plate Now just calculate the height of the electron after it travels4 cm t = .04 / 3.90 * 10E6 = 1.026 * 10E-8 sec Substitute in h = v0y t - 1/2 at2        to getheight of electron as it leaves the plates th = 3.9 * 10E6 / 7.025 * 10E14 = 5.55 * 10E-9sec     time for electron to reach maximumheight tx = .04 / 3.90 * 10E6 = 1.026 * 10E-8sec        greater than time toreach max height hmax = 3.90 10E6 * 5.55 * 10E-9 - 1/2 * 7.025* 10E14 * (5.55 * 10E-9)2 = .011 m = 1.1 cm Thus the electron does not reach the plate The horizontal distance it travels is d = v0x t = 3.9 * 10E6 * 2 * 5.55 * 10E-9 =4.3 cm     the electron does not get back tothe lower plate Now just calculate the height of the electron after it travels4 cm t = .04 / 3.90 * 10E6 = 1.026 * 10E-8 sec Substitute in h = v0y t - 1/2 at2        to getheight of electron as it leaves the plates .

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