A particle of mass m, charge +q approaches a fixed +Q with initial speed u and impact parameter d. Initial and final distances from +Q are very large. Use Qq/(4πε₀mu²d) specified in the diagram to determine closest approach (m). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A particle of mass m, charge +q approaches a fixed +Q with initial speed u and impact parameter d. Initial and final distances from +Q are very large. Use Qq/(4πε₀mu²d) specified in the diagram to determine closest approach (m). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Incoming impact parameter is d; at infinity initial speed is u. Angular momentum around fixed Q is L = mud. Step 2 — At closest approach, radial velocity vanishes. Energy conservation gives (1/2)mu² = L²/(2mrₘᵢₙ²) + (Qq)/(4πε₀rₘᵢₙ). Step 3 — Source gives Qq = 4πε₀mu²d, so electrostatic parameter κ/(mu²) = d. Solve rₘᵢₙ²−2drₘᵢₙ−d² = 0 ⇒ rₘᵢₙ = d(1+√2). Step 4 — d = √2−1 m, hence rₘᵢₙ = (√2−1)(√2+1) = 1 m.