(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An electron enters the lower left side of a parallel plate capacitor and exits precisely at the upper right side (just clearing the upper plate). The initial velocity of the electron is 7*10^6m/s parallel to the plates. The capacitor is 2cm long and its plates are separated by 0.15m. Assume that the electric field is uniform at every point between the plates and find its magnitude.

THE ANSWER IS = 2.09*10^5N/C

Don't mind my weak attempt at diagram.

2. Relevant equations

U(y-direction) = U(x-direction)*tanΦ

F=ma=qE

x=U(x-direction)t

y=U(y-direction)+½at^{2}

3. The attempt at a solution

tanθ=0.15/0.02=7.5

U(x-direction)= 7*10^{6}m/s

U(y-direction)= U(x-direction)*tanΦ= 7.6*10^{6}*7.5=5.25*10^{7}m/s

x=2cm=0.02m, y=0.15m

x=U(x-direction)*t

0.02 = 7*10^{6}*t

t = 2.86*10^{-9}s

y = U(y-direction)+½at

0.15 = 5.25*10^{7}*2.86*10^{-9}+ ½(2.86*10^{-9})^{2}a

0.15=0.15015 + 4.0898*10^{-18}a

a= 3.668*10^{13}ms^{-2}

F = ma = qE

= 9.11*10^{-31}* 3.668*10^{13}= qE

⇒ 1.6*10^{-19}* E = 3.3415*10^{-17}

E=208.85NC^{-1}

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# Homework Help: Calculating the magnitude of an electric field

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