RSS Feeds In the figure, a uniform, upward electric field E of magnitude 5.00x10

Question

RSS Feeds In the figure, a uniform, upward electric field E of magnitude 5.00x103 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 L4 cm and separation d 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity U0 of the electron makes an angle =45 with the lower plate and has a magnitude of 8.99×105 m/s. Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates. 2.94x10-2 m You are correct. Your receipt no. is 164-2855 (f) Len uuu The next electron has an initial velocity which has the same angle =45° with the lower plate and has a magnitude of 6.05×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If net enter the vertical position at which the particle leaveshepacebetween the plates Submit Answer Incorrect. Tries 2/15 Previous Tries

Explanation / Answer

Here acceleration due to electrostatic force a = Eq/m

= 5e3*1.6e-19/9.1e-31 = 8.79e14 m/s^2

Maximum height = [u sin 45 degree]^2/2a = [ 6.05e6*sin 45 degree ]^2/[2*8.79e14]

= 0.0104 m = 1.04 cm

Range = [ u^2 sin 2 theta]/a = [ 6.05e6^2*sin 90 degree ]/[8.79e14] = 0.0416 m = 4.16 cm > 4 cm

It does not strike

For finding vertical position, t = 0.04 / [ 6.05e6 cos 45 degree ] = 9.35e-9 s

using second equation of motion,

vertical position =  [ 6.05e6 sin 45 degree *9.35e-9 - 0.5* 8.79e14*9.35e-9^2]

= 0.00158 m answer

  

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