The drawing shows an electron entering the lower left side of a parallel plate c

Question

The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The Initial speed of the electron is 3.55 times 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume, that the electric field between the plates is uniform everywhere and find its magnitude. What was the acceleration of the electron (magnitude and direction?) ____________ Magnitude of the force of gravity acting on the electron? ___________________ Magnitude of the total force acting on the electron? ___________________ Magnitude of the electric force acting on the electron? ______________________ Magnitude of the electric field within the capacitor? __________

Explanation / Answer

time taken to cross the plates,

t = (2 x 10^-2 m ) / (3.55 x 10^6 m/ s)

t = 5.634 x 10^-9 sec

Perpendicular to the plates,

d = v0 t + a t^2 / 2

0.150 x 10^-2 = 0 + a (5.634 x 10^-9)^2 / 2

a = 9.45 x 10^13 m/s^2 ...........Magnitude of acceleration.

direction = upward

---------------

force of gravity = m g = (9.109x 10^-31)(9.81)

= 8.94 x 10^-30 N

---------------

total force = ma = (9.109 x 10^-31) (9.45 x 10^13)

= 8.61 x 10^-17 N

-----------------------

Fnet = Fe - Fg

Fe = 8.61 x 10^-17 N

-----------------------

Fe = q E

8.61 x 10^-17 = (1.6 x 10^-19) E

E = - 538 N/C

magnitude = 538 N/C , direction = downward

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