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The electric field is a crucial concept when dealing with capacitors. It represents the force per unit charge that exists between the plates of a capacitor. In simpler terms, it's how strong the electric push or pull is in the space between the plates.

For a parallel-plate capacitor, the electric field ( \( E \) ) is calculated using the formula: \( E = \frac{V}{d} \) , where \( V \) is the voltage across the plates, and \( d \) is the distance between them.

In our problem, the electric field should not exceed \( 3.00 \times 10^4 \) V/m. This means that the maximum voltage ( \( V \) ) allowable, for the given separation \( d = 1.50 \, \text{mm} \), is \( 45 \, \text{V} \).

This limit is vital as exceeding the electric field threshold can lead to dielectric breakdown or damage to the capacitor.

The dielectric constant ( \( K \) ) is an important factor that influences the capacity of a capacitor to store charge. It measures how much more electric field can be reduced by the dielectric material compared to the original medium, often air.

When a dielectric material is introduced between the plates of a capacitor, it increases the capacitance by a factor equal to \( K \), modifying the formula of capacitance to \( C = K \times C_0 \), where \( C_0 \) is the original capacitance.

In our exercise, a dielectric material with a dielectric constant of \( 2.70 \) was introduced, increasing the capacitance from \( 8.00 \) pF to \( 21.6 \) pF.

The dielectric constant shows how effective a material is in amplifying a capacitor's ability to hold charge without increasing the voltage.

A parallel-plate capacitor is a simple type of capacitor consisting of two conductive plates separated by a certain distance. This configuration allows it to store electrical energy in an electric field.

The important parameters to consider include:

• Plate Area: Affects how much charge the capacitor can store.
• Plate Separation: The distance between the plates, which influences the electric field strength.
• Dielectric Material: Determines the capacitance and the maximum permissible electric field.

The formula for the capacitance of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_0 \times A}{d} \) when air is the dielectric, where \( \varepsilon_0 \) is the permittivity of free space. With a dielectric, it's \( C = \frac{\varepsilon \times A}{d} \), with \( \varepsilon = K \times \varepsilon_0 \).

This simple structure facilitates understanding of how capacitors store and manage electric charge.

Calculating the charge ( \( Q \) ) that a capacitor can store is pivotal in designing and using electronic circuits. The capacitance ( \( C \) ) and voltage ( \( V \) ) determine the charge stored on the plates of a capacitor, calculated using the formula: \( Q = C \times V \).

In our problem, given \( C_0 = 8.00 \, \text{pF} \) (with air) and \( V = 45 \, \text{V} \), the charge is calculated to be \( 3.60 \times 10^{-10} \) C.

When a dielectric is introduced with \( K = 2.70 \), the capacitance increases to \( 21.6 \times 10^{-12} \, \text{F} \). Keeping the voltage constant, the new maximum charge becomes \( 9.72 \times 10^{-10} \) C.

This demonstrates how the presence of a dielectric enhances a capacitor's ability to store more charge without increasing the voltage.