An electron is released from rest at the negative plate of a parallel plate capa

protisvitfc 2021-11-16 Answered

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
σ=1.8×107C/m2
and the plates are separated by a distance of
1.5×102m.
How fast is the electron moving just before it reaches the positive plate?

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Expert Answer

Crom1970
Answered 2021-11-17 Author has 6 answers
Step 1
Concept
The Electric field due to capacitor having surface charge density σ is
E=σϵ0
The acceleration of electron is given by
a=Fm
a=qEm
We have to find the speed of electron as it reaches the plate
Step 2
The acceleration
a=qσmϵ0
a=(1.6×1019C)(1.8×107Cm2)(9.11×1031kg)(8.85×1012Nm2C2)
a=3.572×1015ms2
Step 3
The velocity
The velocity can be found from the equation of kinematics
v2=v02+2(a)s
v2=0+2(3.572×1015ms2)(1.5×102m)
v2=1.0716×1014m2s2
v=1.0716×1014m2s2
v=1.035×107m/s
Result
v=1.035×107ms
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