Can we change the capacitor capacity by changing existing charge

On the plates when it is connected to the battery?

# [Physics] Can We Change the Capacitance of a Capacitor

capacitancechargedielectricgeometryvoltage

#### Related Solutions

I think you question can be answered succinctly from this point:

Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?

A battery never produces additional charge. Very few things can do this (like a Van de Graaff generator), and even those can only do so locally. A battery pumps charge, and that leaves no room for adding to the total charge count (**c**).

I thought it strange that the problem is starting out saying that the capacitor has on net extra charge, but you have analyzed the situation correctly. Since we're treating it as an infinite parallel plate capacitor, before the battery is connected the charge is concentrated on the outside face of both plates.

When the battery is connected, and the capacitor has fully charged (if you neglect internal resistance this last specifier isn't needed), then you will have additional positive charge on one plate and additional negative charge on the other. Following the eqi-potential principle, these opposite charges will gather on the inner faces of the two plates (**a**).

Now a deeper question: will connecting the battery affect the charge density on the outer faces of the plates? I will argue "no". The important unwavering assumption is a constant electric potential throughout the interior of a plate. Introducing the battery introduces a field in-between the inner surfaces, which results in potential difference equal to the voltage of the battery. You have 4 points (which are infinite planes) along the number line (since this is 1D symmetry) where charge is located, and the field is $E=\sigma/(2 \epsilon_0)$ pointing away from the plane - that is, proportional to the charge density and constant. That means that the field on the outside of the plates won't change when connecting the battery, since the total charge quantity stays the same, the field doesn't diminish with distance from the charges, and you only moved charges around. The field within a plate is zero and the outside charge density remains the same, thus, in order to transition from a zero field to the unchanged outside field strength, you require the same charge density on the outside face (**d**). The potential on the surface is different, but the charge is the same - a very important nuance of electronics.

Thus, options "a", "c", and "d" are both correct.

The solution is incorrect. Capacitance depends on geometry, and a parallel plate with a diagonal connector is plainly a different geometry than a parallel plate so the statement "effective distance between the plates becomes zero" is irrelevant.

Capacitance is not a very useful concept to apply to something that is not a capacitor. If you insist on assigning a capacitance to a simple wire, though, the reasonable value *is* actually infinity. An ideal wire has 0 reactance at all frequencies $\omega$ and reactance is given by $X=\omega L-\frac{1}{\omega C}$. It's clear that this can only be zero for all frequencies for $L=0$ and $C=\infty$.

There is the separate concept of "self capacitance" or "capacitance to ground" which can be computed and is useful in some cases, but it's not infinite in this case and isn't directly related to the capacitance of a parallel plate capacitor to begin with.

## Best Answer

There are indeed voltage dependent capacitors where the capacitance is a function of applied voltage, that is, $C(v)$ where $v$ is applied voltage. Although there may be newer technology these days, there used to be dielectric materials used in capacitors made of ferroelectric materials where the applied voltage polarizes the dipoles in the dielectric thus changing the field strength and thus the capacitance. Usually these are non-linear capacitance changes. More information should be easily located via Google search on voltage dependent capacitors.

The OP question though is not as precise as needed to determine if this is valid information as an answer. In particular, "while connected to a battery" may imply fixed voltage source. Changing charge on existing plates is in fact what happens in capacitors during charging and discharging cycles. With a fully charged capacitor with applied DC voltage, no more current will flow. This means the charge accumulated in the capacitor is now fixed. To change that you change one of the following: (1) voltage, (2) capacitance via changing physical dimensions or insertion of different dielectric material or varying the dielectric material in the capacitor.