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. The charge per unit area on each plate is ? = 2.50 10-7 C/m2, and the plates are separated by a distance of 1.1 10-2 m. How fast is the electron moving just before it reaches the positive plate? _____m/s

Explanation / Answer

=> (1/2) mv^2 = q(sigma/epsilon)?x => v = [(2q(simga/epsilon)?x)/m]^1/2 Now, plugging values, mass of electron, m = 9.10938188 × 10-31 kg and charge of electron, q = 1.60217646 × 10-19 C => velocity, v = [2 * (1.60217646 × 10-19) * (2.50 x 10^-7 / 8.854 x 10-12) * (1.1 x 10^-2 / 9.10938188 × 10-31]^1/2 m/s so v= 10.45 x 10^6 m/s.

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