iPad 11:22 AM 75%-. webassign.net PRACTICE IT Use the worked example above to he

Question

iPad 11:22 AM 75%-. webassign.net PRACTICE IT 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 x 103 N/C pointing in the positive x-direction. (a) Assuming an initial speed of zero, find the speed of a proton at x 0.0800 m with a potential energy of -2.40 x 10. (Asume the potential energy at the point of release is zero.) 1.69536E5 m/s (b) 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·0.190 m, a change in potential energy of 5.05 x 10-17 J. 12 158E6 |m/s EXERCISE HINTS: GETTING STARTED I IMSTUCKI The electron in part (b) travels from x 0.190 m (where it has half the initial speed you previously calculated) to x =-0.160 m within the constant electric field. If there's a change in electric potential energy of-8.41 x 10- J as it goes from x = 0.190 m to x =-0.160 m, find the electron's speed at x =-0.160 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.) 17 m/s Need Help?Read lt Submit Answer Save Progress Save Progress Practice AnotherVersion 4. /9 points I Previous Answers SerCP10 16.AE.006. My Notes Ask Your

Explanation / Answer

Speed of electron at x = 0.190 m is, v1 = [12.158 * 106]/2 = 6.079 * 106 m/s

Let 'v' be speed of electron at x = -0.160 m

Let U1 and U2 be the potential energies of electron at x = 0.190 m and x = -0.160 m respectively.

It is given that [U2 - U1] = -8.41 * 10-17 J

mass of electron, me = 9.109 * 10-31 kg

Then,

mev12/2 + U1 = mev2/2 + U2

=> v2 = 2/me [mev12/2 - (U2 - U1)]

Substituting values, we get,

v2 = 2.216 * 1014

=> v = 1.489 * 107 m/s

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