Find the mechanical work needed to double the separation of the plates of a parallel plate capacitor in vacuo if a cell maintains them at a constant potential difference \(V\) and the area and original separation are \(A\) and \(x\), respectively.

The principle underlying a capacitor's ability to store energy is found in its capacitance. Capacitance, denoted by the symbol \textbf{C}, is a measure of how much electric charge a capacitor can hold per unit of electric potential difference (voltage) across its plates. The formula for the capacitance of a parallel plate capacitor is given by:

\[\begin{equation}C = \frac{\epsilon_0 A}{d}\end{equation}\]
Here, \(\epsilon_0\) represents the permittivity of free space, a constant value which is approximately \(8.85 \times 10^{-12}\) farads per meter (F/m), \(A\) is the area of one of the plates, and \(d\) is the distance between the plates. This relationship illustrates that capacitance is directly proportional to the area of the plates and inversely proportional to the separation distance. Hence, increasing the area or decreasing the distance between the plates will result in a greater capacitance.

In the confines of a parallel plate capacitor, we can calculate the electric field, often denoted by \(E\). The electric field between two plates is uniform and is defined as the force per unit charge. It's calculated using the simplified relation:

\[\begin{equation}E = \frac{V}{d}\end{equation}\]
where \(V\) is the potential difference applied across the capacitor's plates, and \(d\) is the separation between the plates. In terms of charge \(Q\) and the permittivity of free space, the electric field can also be expressed as:

\[\begin{equation}E = \frac{Q}{A \epsilon_0}\end{equation}\]
The significance of this lies in understanding how changing the charge or the distance affects the electric field within a capacitor, a crucial detail when considering the work involved in changing plate separation.

Dealing with the concept of work in the context of capacitors involves understanding how energy is transferred in the system. In our example, to double the separation of the plates while maintaining a constant potential difference, work must be done against the electric force between the plates. The work done can be defined as the product of the force applied to move the plates and the distance over which this force is applied.

\[\begin{equation}W = F \Delta x\end{equation}\]
Here, \(W\) is the work done, \(F\) is the electric force between the plates, and \(\Delta x\) is the change in separation between the plates. In the case of a parallel plate capacitor, as the separation doubles, the work is purely electrical and is equal to the change in the capacitor's energy. Realizing that \(F\) is proportionate to the square of the charge stored (\(Q^2\)), and recalling that the charge remains constant if the potential difference is constant, leads to the understanding that more work is required to separate the plates further.

The electric force acting between the plates of a capacitor is a central concept when it comes to capacitors. This force arises due to the electric field generated by the charges on the plates. The formula to calculate this force is based on Coulomb's law and is given by:

\[\begin{equation}F = \frac{Q^2}{A \epsilon_0}\end{equation}\]
where \(F\) denotes the electric force, \(Q\) the charge on each plate, \(A\) the area of the plates, and \(\epsilon_0\) the permittivity of free space. From this equation, we observe that the force is directly proportional to the square of the charge and inversely proportional to both the area of the plates and the permittivity of the medium. It's important to note that in the scenario of doubling the plate separation, the electric force calculated is used to derive the amount of work needed to achieve the change while the electric potential remains constant.