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 x 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)? What is the magnitude of the force of gravity acting on the electron? What is the magnitude of the total force acting on the electron? What is the magnitude of the electric force acting on the electron? What is the magnitude of the electric field within the capacitor?

Explanation / Answer

Given that

initial speed of the electron u=3.55*10^6 m/s

distance x=2 cm

distance y=0.15 cm

now we find the time taken electron exists from capacitor

time t=x/u=2*10^-2/3.55*10^6=5.634*10^-9 sec

now we find the acceleration of the electron

acceleration of the electron a=2y/t^2=2*1.5*10^-3/(5.634*10^-9)^2=9.5*10^13 m/s^2

the acceleration is +y axis

now we find the magnitude of the electric force

the magnitude of the electric force F=ma=9.11*10^-31*9.5*10^13=86.55*10^-18 N

now we find the magnitude of the electric field

the magnitude of the electric field E=F/q=86.55*10^-18/1.6*10^-19=541 N/c

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