Question

Q5. A uniform electric field of magnitude E 360 V/m is directed in the positive X direction. A proton moves from the origin to the point (x, y)-(30.0 cm, 40.0 cm). (a) Through what potential difference does the charge move? Show your work to score. (b) What is the change in the potential energy of the charge field system? Show your work to score. (c) In this uniform electric field E mentioned above, if an electron was released at rest at the origin, along which direction would it move? (d) What would be its speed, vs, after the electron is released from rest and travels distance d in the uniform E field? Please derive the expression of vy as a function of the given quantities: E, d, e (the elementary charge) and m (the mass of an electron) Show your work to score.

Answer 1

(a) ∆V = -∫E.ds = -∫Ei^.ds = -E∫dx = -E*∆x = -360*0.30 = -108 V

(b) ∆U = q*∆V = -1.6*10^-19*108 = 1.728*10^-17 J

(c) Negative X-direction, as electron moves in opposite direction of Electric Field.

(d) Work done by electron = W = F*d = E*e*d

Then

½mv_f² = KE ( Kinetic Energy )
v_f² = 2KE / m
v_f² = 2*E*e*d/m

v_f = sqrt(2eEd/m)