Question

Four charged particles (A, B, C, D) of mass 'm' and charge 'q' each, are connected by light silk threads of length 'd' forming a tetrahedron floating in outer space. The thread connecting particles A and B suddenly snaps. Find the maximum speed of particle A after that –

• A. $\sqrt{\frac{q^{2}}{8\pi\varepsilon_{0}md}\left\lbrack \frac{3}{\sqrt{3}} - 1 \right\rbrack}$
• B. $\sqrt{\frac{q^{2}}{8\pi\varepsilon_{0}md}\left\lbrack \frac{\sqrt{3}}{3} - 1 \right\rbrack}$
• C. $\sqrt{\frac{q^{2}}{4\pi\varepsilon_{0}md}\left\lbrack \frac{3}{\sqrt{3}} - 1 \right\rbrack}$
• D. $\sqrt{\frac{q^{2}}{4\pi\varepsilon_{0}md}\left\lbrack \frac{\sqrt{3}}{3} - 1 \right\rbrack}$

Answer 1

Answer: (B)

$\sqrt{\frac{q^{2}}{8\pi\varepsilon_{0}md}\left\lbrack \frac{\sqrt{3}}{3} - 1 \right\rbrack}$

Answer 2

After the string between A and B is cut. Particle A and B will pivot downward and particle C and D move upward so that momentum is conserved.

When the particle all lie in the same horizontal plane, then charge in potential energy is greatest so the increment in kinetic energy is greatest. So at this condition particle speed is maximum. Speed of A and B in downward

direction = Speed of C and D in upward direction Charge in Potential energy

DU = DKE = $\left[ \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } \right]$× 4

where v = speed of each particle