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=1.8×10−7C/m2 and the plates are separated by a distance of 1.5×10−2m. How fast is the electron moving just before it reaches the positive plate?

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Crom1970 · Accepted Answer

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→=F→m  a→=qE→m  We have to find the speed of electron as it reaches the plate  Step 2   The acceleration  a→=qσmϵ0  a→=(1.6×10−19C)(1.8×10−7Cm2)(9.11×10−31kg)(8.85×10−12N⋅m2C2)  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)(1.5×10−2m)  v2=1.0716×1014m2s2  v=1.0716×1014m2s2  v=1.035×107m/s  Result   v=1.035×107ms

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

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