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

Question

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 1.5 cm, and the electric field within the capacitor has a magnitude of 2.8 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

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 1.5 cm, and the electric field within the capacitor has a magnitude of 2.8 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Explanation / Answer

assuming a uniform electric field
V/m = J/C/m


2.8E6 J/C/m * 0.015 m * 1.6E-19 C/e- * 1 e- = ______J

the potential energy at the start = the Ke at the end

I get about 6.72E-15 J

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