Discussion
Two particles of mass \(m_{1}\) and \(m_{2}\) and charge \(q_{1}\) and \(q_{2}\) respectively are constrained to move in a straight line along x-axis. The Lagrangian of the system (in esu) is :
Answer: B
The Lagrangian is \(\frac{1}{2} m_{1}\dot{x}_{1}^{2} + \frac{1}{2} m_{2}\dot{x}_{2}^{2}\) \(- q_{1}q_{2}/|x_{1} - x_{2}|\).
The Lagrangian is \(L = T - V\). The kinetic energy of the two particles moving along x is \(T = \frac{1}{2}m_{1}\dot{x}_{1}^{2} + \frac{1}{2}m_{2}\dot{x}_{2}^{2}\). In esu (Gaussian) units the Coulomb potential energy of the pair is \(V = q_{1}q_{2}/|x_{1} - x_{2}|\), with no \(4\pi\epsilon_{0}\) factor. So
\[ L = T - \frac{q_{1}q_{2}}{|x_{1} - x_{2}|} \]
The option with \(+q_{1}q_{2}/|x_{1}-x_{2}|\) is the total energy \(T + V\), the Hamiltonian, not the Lagrangian. The two options with negative kinetic terms correspond to \(V - T\) and \(-(T+V)\), neither of which is the Lagrangian.
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