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 10-27 kg and charge 1.6 10-19 C) are initially held at the corners of a square that is 5.5 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.) vfinal =

Explanation / Answer

When protons are at the corners of a square . System have electric potential energy when they are very far apart

PE of system cab be considered zero.

so from energy conservation initial PE gets converted into KE.

Electric PE of 2 point charge set = kq1q2 / d

there 4 set with a distance apart and 2 set with sqrt(a^2 + a^2) distance apart,.

so initial PE = 4(ke*e/ a) + 2(ke*e / a*sqrt2)

PEi = (4 + sqrt(2)) ke^2 /a

k = 9 x 10^9 , a = 5.5 x 10^-9 m , e = 1.6 x 10^-19 C

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

5.414 x 9 x 10^9 x (1.6 x10^-19)^2 / (5.5 x 10^-9) = 4 x ( 1.7 x 10^-27 x v^2 /2 )

v = 8167.32 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.