Velocity of Electron in Capacitor
Homework Answers
apply ma = Eq
a = 3e6 * 1.6*10^-19/9.11E-31
a = 5.26*10^17 m/s^2
now apply V^2 = 2aS
V^2 = 2* 5.26*10^17 * 0.02
V = 1.45*10^8 m/s
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Velocity of Electron in Capacitor
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An electron is released from rest at the negative plate of a parallel plate capacitor. The...
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An electron is released from rest at the negative plate of a
parallel plate capacitor. The charge per unit area of each plate is
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seperated by a distance of 2.1 x 10-2 m. How fast is the
electron moving just before it reaches the positive plate?
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parallel plate capacitor and accelerates to the positive plate (see
the drawing). The plates are separated by a distance of 1.7 cm, and
the electric field within the capacitor has a magnitude of 2.7 x
106 V/m. What is the kinetic energy of the electron just as it
reaches the positive plate? The figure shows a vertical plate on
the left that is negatively charged and another vertical...
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rest at the negative plate of a parallel plate capacitor. The
charge per unit area on each plate is = 2.0 × 10-7 C/m2, and the
plates are separated by a distance of 1.9 × 10-2 m. How fast is the
electron moving just before it reaches the positive plate? ty for
any help :)
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An electron is released from rest at the negative plate of a
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= 2.0 x 10-7 C/m2, and the plates are
seperated by a distance of 2.1 x 10-2 m. How fast is the
electron moving just before it reaches the positive plate?
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