Effect of charges near a parallel plate capacitor

If I charge a parallel plate capacitor and I insert a charged body near one of the plates will there be any interactions like attraction or repulsion?

What if I disconnect the battery?

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There will be always interaction (even if the capacitor would be uncharged). For the sign of the force take a look at wikipedia. – Fabian Oct 23 '12 at 5:51
Better. Ideally, people should be able to infer from the title on the Phys.SE front page whether they want to click on the post or not. – Qmechanic Oct 23 '12 at 15:37

Consider two plates, with area $a$. Let's say that the left plate has charge $q$ and the right has charge $-q$, so the charge density $\pm\rho = \pm q/a$. The plates are a distance $d$ apart. For parallel plate capacitors, the electric field in the gap is given by:

$E = \rho / \epsilon = q / (\epsilon a)$

The field magnitude is the force per unit charge, and it will be acting from one plate to the other (depending on the charge of the particle). So the particle will feel this force:

$F = E Q = qQ / (\epsilon a)$

Note that as the field has no dependence on the position between the plates, this force is constant. So aligning the plates horizontally we could vary the plate charge $q$ until it cancelled the gravitational force on the particle, and the particle would levitate between the plates (assuming you slowed it down - with no overall force it will continue at whatever speed it has).

The electromagnetic interaction is so much stronger than the gravitational interaction that for charged particles it would be very difficult to balance the forces. On the other hand, macroscopic objects have a charge so nearly neutral that gravity wins in any reasonable field.

The charge on a capacitor is due to the position of the charges, so disconnecting the battery will just stop charging the capacitor. Unless the plates are discharged in some way, the field will persist between them.

This is not free energy, however; the particle itself will have its own effect on the field, so if we added many particles they would all end up on the attracting plate; electrons would land on the positive plate and thus reduce its positive charge until it was entirely discharged, and eventually they would stop when both plates were equally negatively charged. The amount of energy available is simply the electric potential energy, a little less than the energy required to charge it initially.

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If your charged body doesn't have a permanent supply of electric charge,

Of course, there would be electrostatic induction for sure. But, I'd say that it would diminish within certain period of time. As a parallel plate capacitor is always grounded, the induced charges would be neutralized. You can't notice this time though.

The other case: If it is provided by a constant supply of charges,

Now, The interaction would be noticeable. And, it won't be a capacitor afterwards :-)

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