In the figure a uniform, upward electric field of magnitude 2.20 times 10^3 N/C

Question

In the figure a uniform, upward electric field of magnitude 2.20 times 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 = 19.0 cm and separation d = 1.90 cm. An electron is then the shot angle 46.0 degree with the lower plate and has a magnitude of 5.80 times 10^6 m/s (a) Will the electron strike on of the plates? If so, which place, top or bottom? How far horizontally from the left edge will the electron strike?

Explanation / Answer

Here ,

electric field , E = 2.2 *10^3 N/C

L = 19 cm = 0.19 m

theta = 46 degree

u = 5.8 *10^6 m/s

a) acceleration of the electron , a = 2.2 *10^3 * 1.602 *10^-19/(9.11 * 10^-31)

a = 3.87 *10^14 m/s^2

for the maximum flight , time of flight

t = 2 * 5.8 *10^6 * sin(46 degree)/(3.87 *10^14)

t = 2.16 *10^-8 s

maximum vertical height = (5.8 *10^6 * sin(46 degree))^2/(2 * 3.87 *10^14)

maximum vertical height = 0.0225 m = 2.25 cm

hence , the electron will strike a plate

b) it will strike the upper plate

c) for the distance from the left edge

let the time taken is t

0.019 = 5.8 *10^6 * sin(46 degree) * t - 0.5 * 3.87 *10^14 * t^2

solving for t

t = 6.53 *10^-9 s

horizontal distance from left plate = 5.8 *10^6 * cos(46 degree) * 6.53 *10^-9

horizontal distance from left plate = 0.0263 m = 2.63 cm

the horizontal distance from left plate is 2.63 cm

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