Question
In: Physics
An electron is released from rest at the negative plate of a parallel plate capacitor and...
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.4 cm, and the electric field within the capacitor has a magnitude of 2.0 x 10^6 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?
Solutions
Expert Solution
Total force acting on electron= qE
=1.6*10-19*2.0*106
=3.2*10-13 N
Change in KE=work done by field=force*displacement
=3.2*10-13*0.014
=4.48*10-15 J
Thus final KE of electron=4.48*10-15 J or 27.96*103 eV
genius_generous answered 1 year agoRelated Solutions
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