5. C+ -5 points SerCP 10 16.P 019 My Notes Ask Your Teacher A proton is located

Question

5. C+ -5 points SerCP 10 16.P 019 My Notes Ask Your Teacher A proton is located at the origin, and a second proton is located on the x-axis at x1 6.84 fm (1 fm 10 15 m). (a) Calculate the electric potential energy associated with this configuration. (b) An alpha particle (charge 2e, mass 6.64 x 10 27 kg) is now placed at (x2, y2) (3.42, 3.42) fm. Calculate the electric potential energy associated with this configuration. (c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. m/s (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity m/s Need Help? Read It

Explanation / Answer

a)

The potential energy is given by

U = k q1 q2 / r

hence, as q1 = q2 = 1.602E-19 C,

U = 3.37*10^-14 J [answer]

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b)

The potential energy of 3 particles is given by

U = (k q1 q2 / r12) + (k q2 q3 / r23) + (k q1 q3 / r13)

Now, calculating the distances using distance formula,

r12 = 6.84 fm
r23 = 4.8366 fm
r13 = 4.8366 fm

Thus, if q1 = q2 = e, q3 = 2e, then

U = 2.246*10^-13 J [answer]

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c)

Here, we simply subtract parts A and B,

delta U = 1.909*10^-13 J [answer]

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d)

Here, at inifnity, by conservation of energy,

delta U = 1/2 m v^2 = 1.909E-13 J

As m = 6.64E-27 kg,

v = 7.58*10^6 m/s [answer]

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e)

Still, by conservation of energy,

delta U = 2(1/2 m v^2) = 1.909E-13 J

But this time m = 1.67E-27 kg for proton. Thus,

v = 1.07*10^7 m/s [answer]

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