Problem 1 A proton is acted on by a uniform electric field of magnitude 263 N/C

Question

Problem 1 A proton is acted on by a uniform electric field of magnitude 263 N/C pointing in the negative z direction. The particle is initially at rest. (a) In what direction will the charge move? (b) Determine the work done by the electric field when the particle has moved through a distance of 3.85 cm from its initial position (c) Determine the change in electric potential energy of the charged particle (d) Determine the speed of the charged particle. Problem 2 Recall that the "del" operator is defined as the following, in cartesian coor

Explanation / Answer

1. (A) F = q E

so proton will move in the direction of force.

q is +ve so F will have same direction as E.

direction-> negative z direction

(B) delta(V) = - E.d = - 263 x 0.0385 = - 10.13 Volt

Work done = - q deltaV

= (1.602 x 10^-19) (-10.13)

= 1.622 x 10^-18 J

(C) PE = - W = - 1.62 x 10^-18 J

(D) Work done = m v^2 /2 - 0

1.622 x 10^-18 = (1.67 x 10^-27) v^2 / 2

v = 4.41 x 10^4 m/s

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