Use the worked example above to help you solve this problem. A proton is release

Question

Use the worked example above to help you solve this problem. A proton is released from rest at x = -2.00 cm in a constant electric field with magnitude 1.50 times 103 N/C pointing in the positive x-direction. Assuming an initial speed of zero, find the speed of a proton at x = 0.0400 m with a potential energy of -1.44 times 10-17 J. (Assume the potential energy at the point of release is zero.) An electron is now fired in the same direction from the same position. Find the initial speed of the electron (at x = -2.00 cm) given that its speed has fallen by half when it reaches x = 0.240 m, a change in potential energy of 6.25 times 10-17 J. The electron in part (b) travels from x = 0.240 m (where it has half the initial speed you previously calculated) to x = -0.200 m within the constant electric field. If there's a change in electric potential energy of -1.06 times 10-16 J as it goes from x = 0.240 m to x = -0.200 m, find the electron's speed at x = -0.200 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.)

Explanation / Answer

F=qE=ma
we get a=qE/m=(1.6*10^-19)(1500)/1.67*10^-27
we get a=240*10^9 m/s^2
distance moved by proton =0.4+0.02=0.42m
v^2=2*240*10^9 *0.42
we get v=(2*240*10^9 *0.42)

b)F=qE=ma
we get a=qE/m=(1.6*10^-19)(1500)/9.1*10^-31
we get 'a'
distance moved by proton =0.2+0.24=0.44m
v^2=2as
we get v=(2as)

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