a) Two parallel plates are separated by 2.0 cm and have a potential difference b

Question

a) Two parallel plates are separated by 2.0 cm and have a potential difference between them of 30 V. The mass of an electron is 9.1 x 10^-31kg and the charge of an electron is 1.6 x 10^-19 C. The electron field between the plates is constant in magnitude and direction. What is the magnitude of the electric field between the plates?

b) The capacitance of the parallel plates is 3.7 F. What is the magnitude of the charge on each plate?

c) If an electron is initially at rest on the negatively charged plate, and is allowed to accelerate toward the positive plate, what is the speed of the electron just before it hits the positive plate?

Explanation / Answer

a) The distance between the plates, d = 2 cm = 0.02 m The potential difference, v = 30 v So the electric field, E = v/d = 30/0.02 = 1500 volt/m b) Given that The capacitance, C = 3.7 F = 3.7 * 10^-6 F We have a formula for the charge on the capacitor plates is         q = CV = (3.7 * 10^-6)(30) = 111 * 10^-6 = 111 C c) We have a formula as (1/2)mv^2 = eV From the above we have        v = sqrt(2eV/m) = 5.9 * 10^5 m/s

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