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Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conductor, with respect to the outer conductor, is +600 V. An electron is released from rest at the surface of the outer conductor. What is the speed of the electron as it reaches the inner conductor? (e = 1.60 × 10-19 C, m el = 9.11 x 10-31 kg, k = 1/4π ε0 = 8.99 × 109 N · m2/C2)

Respuesta :

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

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