Two identical particles of same mass each m are having same magnitude of charge Q. One…

Physics · JEE Advanced · NTA ExamsElectrostatics

Two identical particles of same mass each m are having same magnitude of charge Q. One particle is initially at rest on a frictionless horizontal plane and the other particle is projected directly towards the first particle from a very large distance with a velocity v. The distance of closest approach of the particle will be
  1. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \mathrm{Q}^{2}}{m v^{2}}\)
  2. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Q}^{2}}{\mathrm{mv} v^{2}}\)
  3. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}^{2}}{\mathrm{~m}^{2} v^{2}}\)
  4. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 Q^{2}}{m^{2} v^{2}}\)

Answer

(A) 1 4 _ 0 4 Q ^ 2 m v^ 2

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