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Arm

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- Homework Statement
- Two positive charges are placed one meter away from eachother. One is held in place and one is allowed to move. Find an equation that models the velocity of the charge allowed to move after it is released from rest.

- Relevant Equations
- $$KE = \frac{mv^2}{2}$$

$$U_E= \frac{k q_1 q_2}{r}$$

$$L = KE - PE$$

$$\frac{\partial L}{\partial x} - \frac{d}{dt} ( \frac{\partial L}{\partial \dot x} ) = 0$$

Here is my epic fail at trying to derive the equation using Lagrange (this was my first time trying to use lagrangian mechanics except for when I memorized the derivation for a pendulum)

$$L = \frac{m \dot r^2}{2} - \frac{k q_1 q_2}{r}$$

$$\frac{\partial L}{\partial r} = \frac{k q_1 q_2}{r^2}$$

$$\frac{\partial L}{\partial \dot r} = m \dot r$$

$$\frac{d}{dt} ( m \dot r ) = m \ddot r$$

$$\frac{k q_1 q_2}{r^2} - m \ddot r = 0$$

$$\ddot r = \frac{k q_1 q_2}{m r^2}$$

$$\dot r = \int \frac{k q_1 q_2}{m r^2}dr = \frac{2 k q_1 q_2}{mr}$$

$$L = \frac{m \dot r^2}{2} - \frac{k q_1 q_2}{r}$$

$$\frac{\partial L}{\partial r} = \frac{k q_1 q_2}{r^2}$$

$$\frac{\partial L}{\partial \dot r} = m \dot r$$

$$\frac{d}{dt} ( m \dot r ) = m \ddot r$$

$$\frac{k q_1 q_2}{r^2} - m \ddot r = 0$$

$$\ddot r = \frac{k q_1 q_2}{m r^2}$$

$$\dot r = \int \frac{k q_1 q_2}{m r^2}dr = \frac{2 k q_1 q_2}{mr}$$

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