A proton is released from rest in a uniform electric field of magnitude 80000 V/

Question

A proton is released from rest in a uniform electric field of magnitude 80000 V/m directed along the positive x axis. The proton undergoes a displacement of 0.2 m in the direction of the electric field as shown in the figure. Find the change in the electric potential if the proton moves from the point A to B. The mass of a proton is 1.672623 times 10-27 kg. 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 0.2 m, starting from rest. Answer in units of m/s

Explanation / Answer

change in electric potential=Ed=80000*0.2=16000 change in potential energy=qEd=(1.6*10^-19)*16000=2.56*10^-15 J 0.5*m*v^2=2.56*10^-15 speed of proton=v=1.75*10^6 m/s

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