Question

Considering the dielectric strength of air, what is the maximum amount of charge that can be stored on the plates of a capacitor that are a distance of \(15 \mathrm{~mm}\) apart and have an area of \(25 \mathrm{~cm}^{2}\) ?

Short answer

Answer: The maximum charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

Step-by-step solution

Convert the given dimensions to meters

First, we need to convert the given dimensions (distance and area) to meters for consistency in calculations: - The distance between the plates is given as \(15 \mathrm{~mm}\). Convert this to meters: \(15 \mathrm{~mm} = 0.015 \mathrm{~m}\). - The area of the plates is given as \(25 \mathrm{~cm}^{2}\). Convert this to square meters: \(25 \mathrm{~cm}^{2} = 2.5 \times 10^{-3} \mathrm{~m}^{2}\).

Calculate the maximum electric field

The dielectric strength of air is given as \(3 \times 10^{6} \mathrm{~V/m}\). This is the maximum electric field that air can withstand without breaking down. We can calculate the maximum electric field in air using: \(E = \frac{V}{d}\) where \(E\) is the electric field, \(V\) is the voltage, and \(d\) is the distance between the plates. We can use the dielectric strength of air to find the maximum voltage, \(V_{max}\), that the capacitor can handle: \(V_{max} = E_{max} \times d\) Substituting the values into the formula, we get: \(V_{max} = (3 \times 10^{6} \mathrm{~V/m}) \times (0.015 \mathrm{~m}) = 45,000 \mathrm{~V}\)

Calculate the capacitance

Using the capacitance formula: \(C = \epsilon_{0} \frac{A}{d}\) where \(\epsilon_{0}\) is the vacuum permittivity constant, \(A\) is the area of the plates, and \(d\) is the distance between the plates. The value of \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m}\). Substitute the values to find the capacitance: \(C = (8.85 \times 10^{-12} \mathrm{~F/m}) \times \frac{2.5 \times 10^{-3} \mathrm{~m}^{2}}{0.015 \mathrm{~m}} = 1.475 \times 10^{-11} \mathrm{~F}\)

Calculate the maximum charge stored

Now that we have the capacitance and the maximum voltage that the capacitor can handle, we can find the maximum charge using: \(Q = C \times V\) where \(Q\) is the charge stored, \(C\) is the capacitance, and \(V\) is the voltage across the plates. Substitute the values: \(Q = (1.475 \times 10^{-11} \mathrm{~F}) \times (45,000 \mathrm{~V}) = 6.6375 \times 10^{-7} \mathrm{~C}\) Therefore, the maximum amount of charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

Key concepts

Dielectric Strength

Dielectric strength is a crucial factor for capacitors, particularly when determining how much electric field they can withstand before breaking down. Think of dielectric strength as the maximum electric field a material can handle without failing or losing its insulating properties. For air, this value is around \(3 \times 10^6 \, \mathrm{V/m}\). When designing or analyzing a capacitor, one must ensure that the electric field does not exceed this threshold, to prevent electrical breakdown that could damage the capacitor.

Electric Field

The electric field, denoted as \(E\), is a region around charged particles or objects where electric forces can be noticed. In our case, it's the effect of the voltage across the plates of the capacitor.
We can calculate the electric field using the formula:

• \(E = \frac{V}{d}\)

Where \(V\) is the voltage and \(d\) is the distance between the plates. For a given dielectric strength of air, the electric field must remain below this maximum to ensure safe capacitor operation.

Capacitance

Capacitance is the measure of a capacitor's ability to store charge. It directly depends on the physical properties of the capacitor, specifically the area of the plates and the distance between them.

• The formula for capacitance is \(C = \epsilon_{0} \frac{A}{d}\)

Where:

• \(C\) is the capacitance.
• \(\epsilon_{0}\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \mathrm{F/m}\).
• \(A\) is the area of the plates.
• \(d\) is the distance between the plates.

By adjusting these factors, one can increase or decrease the capacitance of a capacitor.

Voltage

Voltage across a capacitor determines the electric field within the dielectric medium. It's a crucial parameter as the capacitor's plates must not exceed the maximum voltage that the dielectric material can withstand.
By determining the maximum electric field and the distance between the plates, we can calculate the maximum allowable voltage, ensuring the capacitor operates safely without breakdown.

• The formula used is \(V_{max} = E_{max} \times d\).

Where \(E_{max}\) is the maximum electric field according to the dielectric strength.

Charge Storage

Charge storage in a capacitor is pivotal, as it relates to the amount of electrical energy the capacitor can hold. The maximum charge \(Q\) stored is determined by both capacitance and voltage:

• The formula is \(Q = C \times V\).

Where

• \(Q\) is the charge.
• \(C\) is the capacitance.
• \(V\) is the voltage across the capacitor plates.

Understanding charge storage helps prevent the overloading of capacitors and ensures their efficiency and safety during operation.