What happens when we put a conductor between the plates of a capacitor

What happens when we put a conductor between the plates of a capacitor?

insertion of an otherwise unconnected conductor introduces an isopotential surface. if it is inserted into an existing isopotential plane, it has no effect. otherwise the electric field is changed. Think of it as a change in the geometry of the plate.

if the driver is himself a plate of negligible thickness, you turn a capacitor into a series of two, each with a double capacity (so … it changes noting), otherwise … the two capacitors will have a capacity more than double (since the empty space is less than half) and the overall capacity will be increased.

this would increase the capacity, assuming that the driver does not come into contact with the two bonnet plates .

The farads value of a cap is determined by the spacing between the plates, among other factors. if, say, a cap consists of a plate spaced an inch apart and a piece of 1/4 inch thick conductor is placed between the plates, this spacing would effectively be 3/4 inch. do not consider the trivial case where the driver is in contact with the two plates.

The presence of the driver will effectively reduce the spacing between the plates.

Since the capacitance of a parallel plate capacitor depends on the inverse of plate spacing, the capacitance will increase in a certain way. knowing the geometry of the conductor and the original capacitor, we could determine the effect numerically

if you insert a floating conductor, that is to say: not connected to anything, between the plates, you will get two bigger capacitors in series, which gives you less capacity, depending on the thickness of the driver and its position. When the driver is smaller than the plates, you must calculate the compound value by adding parallel and serial capabilities.

When the inserted conductor is connected to ground, you can have 2 capacitors connected to the ground. their value depends on the position between the plates.

If the conductor is connected with only 1 plate, the capacity will increase, but the breakdown voltage will be decreased.

When the conductor connects to both plates, it stops being a capacitor and starts acting as a low value resistor.

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