An electron, moving at a speed of 7.64X10 6 m/s, enters the region between two p

Question

An electron, moving at a speed of 7.64X106 m/s, enters the region between two parallel plates separated by a distance of 5.22 cm. As it enters, it is moving in a direction parallel to the plates themselves, and is 3.55 cm from the positive plate. Between the plates is a potential difference of 375 V. The mass of the electron is 9.11X10-31 kg.

*Find its velocity in the direction perpendicular to the plane of the plates at the time when it strikes the positive plate & During that time how far will it have traveled in the direction parallel to the plates?

Explanation / Answer

electric field between plates, E = V/d

= 375/0.0522

= 7184 N/c

let a is the accelration of electron towards positive plate

q*E = m*a

==> a = q*E/m

= 1.6*10^-19*7184/(9.1*10^-31)

= 1.26*10^15 m/s^2

now Apply

v^2 - u^2 = 2*a*d

v = sqrt(2*a*d) ( in perpendicular direction initial velosity is zero)

v = sqrt(2*1.26*10^15*0.0355)

= 9.46*10^6 m/s <<<<<<<<<---------Answer

time taken, d = 0.5*a*t^2

t = sqrt(2*d/a)

= sqrt(2*0.0355/(1.5*10^15))

= 6.88*10^-9 s

x = vx*t

= 7.64*10^6*6.88*10^-9

= 0.0526 m

= 5.26 cm <<<<<<<<<<--------------Answer

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