# Capacitance of a capacitor

Why does the capacitance of a capacitor increase if the distance between the two plates of a parallel plate capacitor is decreased? I think, with decreasing distance between the two plates, the force of attraction between the charges on the two plates will increase, and as a result more charge will be stored.

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Consider a charge-free capacitor connected to a battery in an RC circuit.

When the circuit is switched on, the capacitor will begin to accumulate charge. When will it stop? Have a look at the top capacitor plate in the diagram. As the top plate accumulates more and more charge, charge carriers coming in through the top wire will be increasingly repelled by the accumulating charge on the top plate. However, at the same time, the same charges being repelled will also be "pulled" to the top plate by the buildup of opposite charge on the bottom plate. When the plates are close together, the incoming charges moving toward the top plate will feel a stronger pull toward the bottom plate, because the opposite charges on the bottom plate will be closer to the incoming charges on the top. If the plates are far apart, then the "pull" from the other plate won't be very strong, and only a small amount of charge will have to build up on the top plate before the local repulsion dominates and the plate gets "full". Because, in the close-plated scenario, the incoming charges experience a greater "pull" than they do when the plates are far apart, more of them can build up on the plate before the capacitor fills up.

Hope this helps.

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as $C=E_0\cdot A / d$, the nearness of plates increases the capacitance.

if d decrease, capacitance $C$ increases.

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A capacitor has a capacitance depending on the maximum voltage sustained across the plates $C_{\text{max}}=\frac Q{\Delta V_{\text{max}}}$.

The voltage sustained between the plates depends on field strength between the plates. The strength of the electric field between the plates is equal to the voltage sustained between the plates divided by the distance between the plates $E=\frac {\Delta V}{\Delta x}$.

The strength of the electric field is limited by Electric Breakdown. Electric breakdown occurs when the strength of the field between the plates exceeds the strength of the dielectric between the plates.

Under a high enough electric field the dielectric partially ionises. The ions in the field accelerate and collide with other molecules in the dielectric creating more ions. The ions form conducting channels between the plates which allows current to flow in spontaneously discharge i.e. sparks/lightning between the plates .

Combining these concepts we see;$$C_{\text{max}}=\dfrac {Q\Delta x}{E_{\text{max}}}$$ the capacitance is proportional to the distance between the plates $C_{\text{max}}\propto \Delta x$.

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1) Capacitance is the ratio of the stored charge Q (+ on either plate, - on the other) to the voltage between the plates.

2) The electric field between the plate is proportional to the plate surface charge density $\sigma = Q/A$, with $A$ the plate area.

3) So if one moves the plates closer together while holding the charge fixed (no external circuit connections), a) the electric field also stays fixed (by 2 above), b) the voltage (= electric field x spacing) decreases because the spacing is decreasing, and c) the capacitance ($C=Q/V$) increases.

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