Calculate the force of attraction \(F\) bet ween the plates of a parallel-plate capacitor. Assume that the capacitor is connected to a battery supplying a constant voltage \(V\). Use the method of virtual work, assuming a small increase \(d s\) in the spacing \(s\) between the plates, and set $$ d W_{B}+d W_{m}=d W_{e} $$ where \(d W_{B}\) is the work done by the battery, \(d W_{m}=F d s\) is the mechanical work done on the system, and \(d W_{e}\) is the increase in the energy stored in the electric field. You should find that $$ F=\frac{1}{2} \epsilon_{0} \frac{V^{2}}{s^{2}} S=\frac{1}{2} \epsilon_{0} E^{2} S $$ Note that one half of the energy supplied by the battery appears as mechanical work, and one half as an increase in the energy stored in the electric field. This is a general rule.

Electrostatic force is a fundamental aspect of electricity and magnetism, particularly relevant when discussing the behavior of parallel-plate capacitors. This force arises due to the electric charges on the plates, attracting or repelling each other. In our problem, the plates are oppositely charged, resulting in an attractive force.

The force of attraction between two plates can be calculated using the formula derived from the method of virtual work, showing that the electrostatic force is proportional to the square of the voltage applied and inversely proportional to the square of the separation between plates. To understand the exercise better, recall that electrostatic force is the underlying principle making capacitors functional: they store energy due to the separation of charge.

It's essential to visualize the capacitor plates with charges on them: positive charge accumulates on one plate and negative charge on the other. The electrostatic force between these charges is what we calculate when we seek the force of attraction in a parallel-plate capacitor. Additionally, this force is closely linked to the concept of electric field strength and the energy stored in the capacitor.

A capacitor's ability to store energy is one of its most valuable features. The energy stored in a capacitor comes from the work required to separate positive and negative charges against the electrostatic force. This stored energy, or potential energy, is what powers circuits in the absence of a continuous power supply.

The energy stored in a parallel-plate capacitor can be expressed as \( U = \frac{1}{2}CV^2 \), where \( C \) represents capacitance and \( V \) the voltage. Capacitance, in turn, depends on the physical characteristics of the capacitor, such as the area of the plates, \( S \), and their separation, \( s \), along with the permittivity of free space, \( \epsilon_0 \).

The step-by-step solution to our problem reveals that when the separation between the capacitor plates increases, it involves work done against the electrostatic force. This work is manifested as an increase in the energy stored within the electric field between the plates. Recognizing this relationship not only assists in solving problems involving force calculations but also underscores the significance of considering energy changes in electrical systems.

The method of virtual work is a powerful principle in physics and engineering used to compute forces and energies within a system. It involves envisioning a 'virtual' infinitesimal displacement or change in the system and calculating the work done during this hypothetical process.

In our exercise, this concept helps calculate the force of attraction between the plates of a capacitor by considering a small increase in plate separation \( d s \). This virtual displacement doesn't actually occur but allows us to determine the relationship between energy input, output, and storage. Via this method, we find that when the plate separation increases, the battery does work to maintain the voltage, while at the same time mechanical work is done against the electrostatic force. The sum of these works equals the increase in the stored energy.

Understanding the method of virtual work betters problem-solving skills by allowing students to conceptualize energy transformations in a system. Its application in the solution to our problem is a great example of how energy considerations can lead to a deeper understanding of physical forces in electrostatic systems.

Electric field strength, denoted by \( E \), is a measure of the force per unit charge exerted by an electric field over a charged object. In the context of a parallel-plate capacitor, the electric field is uniform and directed from the positive to the negative plate, with a strength calculated by \( E = \frac{V}{s} \), where \( V \) is the voltage between the plates and \( s \) is their separation.

This field strength is crucial for understanding the behavior of capacitors. It tells us how strong the force on a charge will be if placed within the field; consequently, it also relates to the force between the plates of the capacitor itself.

In our exercise, we see that the electrostatic force of attraction between the capacitor plates can be expressed in terms of the electric field strength: \( F= \frac{1}{2} \epsilon_0 E^2 S \). This provides a direct illustration of how the field strength, resulting from the applied voltage, dictates the magnitude of the electrostatic force. Grasping electric field strength allows students to appreciate the ability of capacitors to create forces, store energy, and power electrical devices.