This is a challenging multi-step problem. Solve it on paper, writing out each st

Question

This is a challenging multi-step problem. Solve it on paper, writing out each step carefully. When doing calculations, do not round intermediate values. Note: If you have approached the problem in a principled way, do not abandon your approach if your numerical answer is not accepted; check your calculations and make sure you have entered your answer correctly. Four protons (each with mass 1.7 x 10'27 kg and charge 1.6 x 10-19 C) are initially held at the corners of a square that is 6.7 x 10'9 m on a side. They are then released from rest. What is the speed of each proton when the protons are very far apart? (You may assume that the final speed of each proton is small compared to the speed of light.) '"'final = m/s

Explanation / Answer

Electric potential energy of 2 charge system = kq1q2 / d

so there are total 6 sets of proton.

4 idential set S1 = proton with distance between them equal to side of square

2 identical set S2 = proton with distance between them equal to diagonal length

for S1 :

energy of 1 set E = kee / L = ke^2 / L

where e = charge on proton = 1.6 x 10^-19 J

L = side of square

total PE of S1 = 4E = 4ke^2 / L

for s2:

d = sqrt(L^2 + L^2) = sqrt(2) L

PE of 1 set = ke^2 / sqrt(2) L

total PE os set = 3E = 2 ke^2 / sqrt(2) L = sqrt(2) ke^2 / L

initial total PE = 4ke^2 / L + sqrt(2) ke^2 / L

as protons goes to infinity, PE will become zero and it will convert into KE>

so , 4ke^2 / L + sqrt(2) ke^2 / L = 4mv^2 /2

(4 + sqrt(2)) ke^2 / L = 2mv^2

m = mass of proton = 1.673 x 10^-27 kg

putting all values,

v = 7459.48 m/s

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