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 7.10 106 m/s. The capacitor is x = 2.25 cm long, and its plates are separated by y = 0.193 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. --------- N/C

7. 0/1 points I Previous Answers CJ9 18.P052 My Notes Ask Your Teacher 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 7.10 x 10° m/s. The capacitor is x 2.25 cm long, and its plates are separated by y = 0.193 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude 354.29x N/C

Explanation / Answer

We need electric field between the plates, Using force balance on elecctron

Fe = Fnet

from newtorn's 2nd law Fnet = me*a

Force on electron due to electric field = Fe = q*E

where, q = charge on electron = e = 1.6*10^-19

me = mass of electron = 9.1*10^-31

So

e*E = me*a

E = me*a/e

Now to find the acceleration of electron, Using 2nd kinematic equation in y-direction

y = Uy*t + (1/2)*ay*t^2

Uy = initial speed of electron = 0 m/sec

y = plate separation distance = 0.193 cm = 0.193*10^-2 m

t = time taken

a = 2y/t^2

for time using same equation in x-direction

x = Ux*t + (1/2)*ax*t^2

Ux = initial speed in x-axis = 7.1*10^6 m/sec

ax = acceleration in x-direction = 0

x = length of capacitor = 2.25 cm

t = x/Ux = 2.25*10^-2/(7.1*10^6)

t = 3.17*10^-9 sec

Now Using this t

ay = a = 2*0.193*10^-2/(3.17*10^-9)^2

a = 3.84*10^14 m/sec^2

So using this value of a

E = me*a/e

E = 9.1*10^-31*3.84*10^14/(1.6*10^-19)

E = 2184 N/C

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