Consider a parallel-plate capacitor having an area of \(3225 \mathrm{mm}^{2}\left(5 \text { in. }^{2}\right),\) a plate separation of \(1 \mathrm{mm}(0.04 \text { in. })\), and with a material having a dielectric constant of 3.5 positioned between the plates. (a) What is the capacitance of this capacitor? (b) Compute the electric field that must be applied for \(2 \times 10^{-8} \mathrm{C}\) to be stored on each plate.

A parallel-plate capacitor consists of two conducting plates separated by a small distance with an insulating material between them, known as a dielectric. The capacity of this device to store electric charge is measured in farads, and is termed as its capacitance. The formula to calculate the capacitance of a parallel-plate capacitor is given by:

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In the example provided, the area, plate separation, and dielectric constant are specified, which allows us to calculate the capacitance by applying the given values into the standard formula.

The dielectric constant, also known as relative permittivity (\(\varepsilon_r\)), is a measure of a material's ability to resist the formation of an electric field within it. Simply put, it is the ratio of the permittivity of a material to the permittivity of free space (\(\varepsilon_0\)).

• The higher the dielectric constant, the better the material acts as an insulator.
• A dielectric constant of 1 indicates a vacuum or free space.
• Different materials have different dielectric constants, which can affect the overall capacitance of capacitors in which they are used.

The incorporation of a dielectric material between capacitor plates increases the capacitance as compared to that of a vacuum. This occurs because the dielectric reduces the electric field and allows the capacitor to store more charge for the same voltage. For the given exercise, the dielectric constant is 3.5, which enhances the capacitance of the parallel-plate capacitor beyond what it would be if only vacuum were present between the plates.

Electric field strength (\(E\)) represents the force per unit charge experienced by a small positive test charge placed within the field. Its unit is volts per meter (\(\text{V/m}\)).

• The electric field strength between the plates of a capacitor is directly related to the voltage applied across the plates and inversely related to the distance between them.
• Mathematically, it is given by the equation \(E = \frac{V}{d}\), where \(V\) is the voltage and \(d\) is the separation between plates.

The significance of electric field strength within a capacitor setup lies in its ability to determine how much charge the capacitor can store at a given voltage. A stronger electric field, for a constant voltage, implies that the plates can hold more charge. In our particular problem, after computing the voltage needed for a charge storage of \(2 \times 10^{-8} \, \text{C}\), the electric field strength is calculated using the plate separation and found to be \(1.9324 \times 10^5 \, \text{V/m}\). This tells us the intensity of the electric field necessary to maintain the specified charge on the plates of the given capacitor.