Consider a parallel-plate capacitor having an area of \(2500 \mathrm{~mm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\), and with a material of dielectric constant \(4.0\) positioned between the plates. (a) What is the capacitance of this capacitor? (b) Compute the electric field that must be applied for \(8.0 \times 10^{-9} \mathrm{C}\) to be stored on each plate.

Understanding capacitance is essential when studying parallel-plate capacitors. Capacitance, measured in farads (F), is a measure of a capacitor’s ability to store an electric charge. For a parallel-plate capacitor, the capacitance can be calculated using the formula \(C = \frac{\kappa \epsilon_0 A}{d}\), where \(C\) is the capacitance, \(\kappa\) is the dielectric constant, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the separation between the plates.

To make this concept more approachable, let's apply this formula to an example. A capacitor has plates with an area of \(2500 \mathrm{~mm}^{2}\) and a separation of \(2 \mathrm{~mm}\). A dielectric material with a constant value of \(4.0\) fills the space between the plates. Inserting these values yields a capacitance of \(88.5 \mathrm{pF}\), which indicates the capacitor's storage capacity.

A dielectric constant, often denoted by \(\kappa\), is a number that describes how a dielectric material affects the electric field and the capacitance of a capacitor. A dielectric is an insulating material placed between the plates of a capacitor and has two important effects. Firstly, it increases the amount of charge the capacitor can store by reducing the electric field inside the material.

Implications of Dielectric Constant

In our example, the dielectric constant is \(4.0\). It directly multiplies the capacitance by the same factor, compared to the capacitance in a vacuum, which is represented by the vacuum permittivity \(\epsilon_0\). This means that with the dielectric in place, the capacitor can store four times as much charge as it would without the dielectric material, for a given voltage across its plates.

Electric field strength, denoted as \(E\), describes the force per unit charge experienced by a stationary test charge in an electric field. It's measured in volts per meter (V/m). The electric field in a capacitor is uniform between the plates and can be calculated using the formula \(E = \frac{V}{d}\), where \(V\) is the voltage across the capacitor's plates, and \(d\) is the distance between them.

Understanding Applied Electric Fields

For our example capacitor, to store \(8.0 \times 10^{-9} \mathrm{C}\) of charge, a voltage of \(9.04 \times 10^{2} \mathrm{V}\) is needed across the plates. This voltage creates an electric field strength of \(4.52 \times 10^{5} \mathrm{V/m}\). This field strength represents the force that would act on charges within the capacitor, influencing how the charge is stored.