Four identical point charges Q are located at the four corners of a square of si

Question

Four identical point charges Q are located at the four corners of a square of side a. Show that the electric potential energy (EPE) of the system is given by U = (4 + squareroot 2) k_cQ^2/a When the nucleus of uranium-235 (92 protons and 143 neutrons) absorbs an additional neutron, it undergoes a process called nuclear fission in which it breaks into two smaller nuclei. One possible fission is for uranium nucleus to divide into two palladium nuclei, each of which has 46 protons and 5.9 times 10^-15 m in radius. If (immediately after the fission) we assume that the palladium nuclei are at rest and are just touching each other, what is the electric potential energy of the system in electron volts?

Explanation / Answer

1) let q1 = q2 = q3 = q4 = Q

Total potential energy = k*q1*q2/a + k*q1*q3/(sqrt(2)*a) + k*q1*q4/a + k*q2*q3/a + k*q2*q4/(sqrt(2)*a) + k*q3*q4/a

= k*Q^2/a + k*Q^2/(sqrt(2)*a) + k*Q^2/a + k*Q^2/a + k*Q^2/sqrt(2)*a) + k*Q^2/a

= 4*k*Q^2/a + 2*k*Q^2/(sqrt(2)*a)

= 4*k*Q^2/a + sqrt(2)*k*Q^2/a

= (4 + sqrt(2))*k*Q^2/a

2)

let q1 = q2 = Q = 46*1.6*10^-19

d = 2*5.9*10^-15 m

Potential energy of the two nucie, U = k*q1*q2/d

= k*Q^2/d

= 9*10^9*(46*1.6*10^-19)^2/(2*5.9*10^-15)

= 4.13*10^-11 J

= 4.13*10^-11/(1.6*10^-19)

= 2.58*10^8 eV

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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