A proton is released from rest in a uniform electric field of magnitude 1.7 time

Question

A proton is released from rest in a uniform electric field of magnitude 1.7 times 105 V / m directed along the positive x axis. The proton undergoes a displacement of 0.4 m in the direction of the electric field as shown in the figure. The mass of a proton is 1.672623 times 10 - 27 kg. Find the change in the electric potential if the proton moves from the point A to B. Answer in units of V. Find the change in potential energy of the proton for this displacement. Answer in units of J. Apply the principle of energy conservation to find the speed of the proton after it has moved 0.4 m, starting from rest. Answer in units of m / s.

Explanation / Answer

not too sure about this one. But I used energy or conservation. Sopretty sure its right. .5mv^2=U .5mv^2=1.088 x 10^-14 .5(1.672623x10^-27)v^2=1.088x10^-14 v^2=1.3x10^13 v = 3.607 x 10^6 m/s

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