In the figure, a uniform, upward electric field $\\vec{E}$ of magnitude 4.50?103

Question

In the figure, a uniform, upward electric field $ec{E}$ of magnitude 4.50?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 = 4 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 $ec{v}$0 of the electron makes an angle ?=45? with the lower plate and has a magnitude of 9.64?106 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.

A) answer is 2.56 cm

need help with B

The next electron has an initial velocity which has the same angle ?=45? with the lower plate and has a magnitude of 7.25?106 m/s. Will this 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.

I got 0.0184 m but it says I'm incorrect

Explanation / Answer

b) acceleration = a = F/m = eE/m = -(1.602*10^-19*4.50×10^3)/(9.109*10^-31) = -7.91415*10^14m/sec^2 vertical velocity = 7.25×10^6*sin45 = 5.126524164*10^6 at maximum height vertical velocity = 0 v^2-u^2 = 2as =>u^2 = -2as =>s = -u^2/2a = 0.016603961m = 1.66cm < 2cm....So it will not strike. Let it leaves at time t. No acceleration in horizontal direction x = ut =>t = x/u u = horizontal velocity = 7.25×10^6*cos45=5.126524164*10^6 x = 4cm = 0.04 =>t = 7.80256*10^-9 seconds Now in vertical direction: s = ut+0.5at^2 =>s = (5.126524164*10^6*7.80256*10^-9)-(0.5*7.91415*10^14*7.80256*10^-9*7.80256*10^-9) = 0.016m = 1.6cm

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