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But what if both charges are allowed to move? Is the original potential energy shared between the two charges? Would you use half of the original potential energy for each of the two charges?

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- Thread starter UMath1
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- #1

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But what if both charges are allowed to move? Is the original potential energy shared between the two charges? Would you use half of the original potential energy for each of the two charges?

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Simon Bridge

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You can work out the potential energy between the initial and final positions - this will tell you the total change in kinetic energy.

The symmetry dictates how the kinetic energy is shared.

You can also work the problem by brute force using Lagrangian mechanics.

A lot depends on what, exactly, you are trying to calculate.

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Khashishi

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Simon Bridge

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For N particles, positions ##\vec r_i## Newton's second law for the jth particle is $$k\sum_{i\neq j}^N \frac{q_iq_j}{|\vec r_j-\vec r_i|^3}(\vec r_j-\vec r_i) = m_j\frac{d^2}{dt^2}(\vec r_j-\vec r_i)$$

... the result is a system of N differential equations which you can solve for each ##r_i(t)## by the usual methods.

For the case of two identical particles, ##q_1=q_2=q## and ##m_1=m_2=m## this simplifies to:

$$\frac{kq^2}{|\vec r_2-\vec r_1|^3}(\vec r_2-\vec r_1) = m\frac{d^2}{dt^2}(\vec r_2-\vec r_1)\\

\frac{kq^2}{|\vec r_1-\vec r_2|^3}(\vec r_1-\vec r_2) = m\frac{d^2}{dt^2}(\vec r_1-\vec r_2)\\

\vec r_1(0)=\vec r_{01}, \vec r_2(0)=\vec r_{02}, \frac{d}{dt} \vec r_1(0) = \frac{d}{dt} \vec r_2(0) = 0$$

It is usually easier to change coordinates so that one axis (usually z) lies along ##\vec r_2-\vec r_1## with the origin half way between the particles.

This way the particles start at ##\pm z_0## given by ##z_0 = \frac{1}{2}|\vec r_1-\vec r_2|##.

The equations become:

$$\frac{kq^2}{(z_2-z_1)^2}\hat k = m\frac{d^2z_2}{dt^2}\hat k\\

\frac{kq^2}{(z_2-z_1)^2}\hat k = -m\frac{d^2z_1}{dt^2}\hat k\\

z_1(0)=-z_0, z_2(0)=z_0, \dot z_1=\dot z_2 = 0$$ ... something like that.

But if the particles start out stationary at ##\pm z_0## at some time, and they are later at ##\pm z_1## so that ##z_1>z_0##, then, ceteris paribus, we can work out their velocities by conservation of energy:

$$v^2 = \frac{kq^2}{m}\left(\frac{1}{z_0}-\frac{1}{z_1}\right)\\ \vec v_2=v\hat k = -\vec v_1$$

... the result is a system of N differential equations which you can solve for each ##r_i(t)## by the usual methods.

For the case of two identical particles, ##q_1=q_2=q## and ##m_1=m_2=m## this simplifies to:

$$\frac{kq^2}{|\vec r_2-\vec r_1|^3}(\vec r_2-\vec r_1) = m\frac{d^2}{dt^2}(\vec r_2-\vec r_1)\\

\frac{kq^2}{|\vec r_1-\vec r_2|^3}(\vec r_1-\vec r_2) = m\frac{d^2}{dt^2}(\vec r_1-\vec r_2)\\

\vec r_1(0)=\vec r_{01}, \vec r_2(0)=\vec r_{02}, \frac{d}{dt} \vec r_1(0) = \frac{d}{dt} \vec r_2(0) = 0$$

It is usually easier to change coordinates so that one axis (usually z) lies along ##\vec r_2-\vec r_1## with the origin half way between the particles.

This way the particles start at ##\pm z_0## given by ##z_0 = \frac{1}{2}|\vec r_1-\vec r_2|##.

The equations become:

$$\frac{kq^2}{(z_2-z_1)^2}\hat k = m\frac{d^2z_2}{dt^2}\hat k\\

\frac{kq^2}{(z_2-z_1)^2}\hat k = -m\frac{d^2z_1}{dt^2}\hat k\\

z_1(0)=-z_0, z_2(0)=z_0, \dot z_1=\dot z_2 = 0$$ ... something like that.

But if the particles start out stationary at ##\pm z_0## at some time, and they are later at ##\pm z_1## so that ##z_1>z_0##, then, ceteris paribus, we can work out their velocities by conservation of energy:

$$v^2 = \frac{kq^2}{m}\left(\frac{1}{z_0}-\frac{1}{z_1}\right)\\ \vec v_2=v\hat k = -\vec v_1$$

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