A proton with a speed of 1.75 middot 10^4 m/s is moving from infinity directly t

Question

A proton with a speed of 1.75 middot 10^4 m/s is moving from infinity directly toward a alpha particle. Assuming that the alpha particle (with charge of 2e) fixed in place, find the position where the moving proton stops momentarily before turning around. A positive charge of 4.50 mu C is fixed in place. A particle of mass 5.00 g and charge +3.00 mu C is fired with an initial speed of 55.0 m/s directly toward the fixed charge from distance of 4.20 cm. How close does the moving charge get to the fixed charge?

Explanation / Answer

here,

Use conservation of energy (K + U) = constant where U = k*q1*q2/r

So (K + U)1 = (K + U)2

now U1 = 0 and K2 = 0

so 1/2*mp*v^2 = k*q*q/r

1/2*mp*v^2 = k*e*2*e/r

so 0.5 * 1.67 * 10^-27 * (1.75 * 10^4)^2 = 9 * 10^9 * 2 *(1.6 * 10^-19)^2/r

solving for r

r = 1.8 * 10^-9 m

the distance of closest approach is 1.8 * 10^-9 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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