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Consider the system below comprising a pair of oppositely charged parallel plates, connected by a rigid rod of length [itex]r[/itex], constrained to move in the x-direction.

The forces on the plates from the Coulomb electric fields and the magnetic forces are all in the vertical direction so that they do not affect the motion of the system.

The only fields that do affect the system are the horizontal radiation fields due to the acceleration of the charged plates.

Let us assume that we initially apply a force [itex]F_0[/itex] so that we have:

[tex]F_0 = m a_0[/tex]

As charge [itex]+q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the negative plate given by:

[tex]E_+ = \frac{-qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

This will lead to a horizontal force [itex]f_0[/itex] on the negative plate, pointing towards the right, given by:

[tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Likewise as charge [itex]-q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the positive plate given by:

[tex]E_- = \frac{qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Again this will lead to a horizontal force [itex]f_0[/itex] on the positive plate, pointing towards the right, given by:

[tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Thus the total force on the system is now given by:

[tex]F_1 = F_0 + 2f_0[/tex]

Therefore the system has an increased acceleration given by:

[tex]a_1 = \frac{F_1}{m}[/tex]

[tex]a_1 = a_0 \left(1+\frac{q^2}{2\pi\epsilon_0c^2rm}\right)[/tex]

The argument can be repeated indefinitely leading to the calculated acceleration diverging exponentially.

What's gone wrong?

There are no [itex]\dot{a}[/itex] terms in the calculation so I don't think the Abraham-Lorentz reaction force due to the production of electromagnetic radiation can save the day. A positive feedback involving forces and accelerations seems to preclude a consistent initial value for the acceleration.

Here's my (rather radical) solution: perhaps the mass has to change to ensure that the acceleration stays constant.

Perhaps we have:

[tex]F_0 = m_0 a_0[/tex]

[tex]F_1 = m_1 a_0[/tex]

Therefore:

[tex]\frac{m_1}{m_0} = \frac{F_1}{F_0}[/tex]

[tex] \frac{m_1}{m_0} = \frac{1}{m_0a_0}\left(m_0a_0+\frac{q^2a_0}{2\pi\epsilon_0c^2r}\right)[/tex]

[tex]m_1 = m_0 + \frac{q^2}{2\pi\epsilon_0c^2r}[/tex]

The forces on the plates from the Coulomb electric fields and the magnetic forces are all in the vertical direction so that they do not affect the motion of the system.

The only fields that do affect the system are the horizontal radiation fields due to the acceleration of the charged plates.

Let us assume that we initially apply a force [itex]F_0[/itex] so that we have:

[tex]F_0 = m a_0[/tex]

As charge [itex]+q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the negative plate given by:

[tex]E_+ = \frac{-qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

This will lead to a horizontal force [itex]f_0[/itex] on the negative plate, pointing towards the right, given by:

[tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Likewise as charge [itex]-q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the positive plate given by:

[tex]E_- = \frac{qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Again this will lead to a horizontal force [itex]f_0[/itex] on the positive plate, pointing towards the right, given by:

[tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

Thus the total force on the system is now given by:

[tex]F_1 = F_0 + 2f_0[/tex]

Therefore the system has an increased acceleration given by:

[tex]a_1 = \frac{F_1}{m}[/tex]

[tex]a_1 = a_0 \left(1+\frac{q^2}{2\pi\epsilon_0c^2rm}\right)[/tex]

The argument can be repeated indefinitely leading to the calculated acceleration diverging exponentially.

What's gone wrong?

There are no [itex]\dot{a}[/itex] terms in the calculation so I don't think the Abraham-Lorentz reaction force due to the production of electromagnetic radiation can save the day. A positive feedback involving forces and accelerations seems to preclude a consistent initial value for the acceleration.

Here's my (rather radical) solution: perhaps the mass has to change to ensure that the acceleration stays constant.

Perhaps we have:

[tex]F_0 = m_0 a_0[/tex]

[tex]F_1 = m_1 a_0[/tex]

Therefore:

[tex]\frac{m_1}{m_0} = \frac{F_1}{F_0}[/tex]

[tex] \frac{m_1}{m_0} = \frac{1}{m_0a_0}\left(m_0a_0+\frac{q^2a_0}{2\pi\epsilon_0c^2r}\right)[/tex]

[tex]m_1 = m_0 + \frac{q^2}{2\pi\epsilon_0c^2r}[/tex]

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