A uniform electric field of magnitude 410 N/C pointing in the positive x -direct

Question

A uniform electric field of magnitude 410 N/C pointing in the positivex-direction acts on an electron, which is initially atrest. The electron has moved 3.10 cm. (a) What is the work done by the field on theelectron?
1 J

(b) What is the change in potential energy associated with theelectron?
2 J

(c) What is the velocity of the electron?
magnitude 3 m/s direction 4---Select----x-y+y+x (a) What is the work done by the field on theelectron?
1 J

(b) What is the change in potential energy associated with theelectron?
2 J

(c) What is the velocity of the electron?
magnitude 3 m/s direction 4---Select----x-y+y+x magnitude 3 m/s direction 4---Select----x-y+y+x

Explanation / Answer

(a). work done W = - V q                             = - E d q where E = electric field = 410 N / C            q= charge of electron = 1.6 * 10 ^ -19 C            d= separation of the plates = 3.1 cm = 0.031 m substitue values we get W = - 20.336 * 10 ^ -19 J (b). the change in potential energy associated with theelectron is P = -W                                                                                                     = 20.336 * 10 ^ -19 J              (c). Velocity of the electron v = [ 2V q / m]                                              = [ 2Edq / m ] where m = mass of electron = 9.1 * 10 ^ -31 kg substitue values we get v = 6.68 * 10 ^ 7 m / s

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