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 1.0 cm, and the electric field within the capacitor has a magnitude of 2.5 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 1.0 cm, and the electric field within the capacitor has a magnitude of 2.5 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Explanation / Answer

KE = eV

KE = eEd   = 1.6*10^-19 * 2.5*10^6 * 0.01

KE = 4*10^-15 J

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