Question

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

Short answer

(a) 3.60 × 10⁻¹⁰ C; (b) 9.72 × 10⁻¹⁰ C with dielectric.

Step-by-step solution

Understanding the Problem

We need to determine the maximum charge that can be placed on the capacitor's plates without exceeding a certain electric field (E) limit, firstly when there's only air between the plates and secondly when a dielectric material is inserted. The formulas we'll use involve the relationships between capacitance, charge, voltage, and dielectric materials.

Calculate Max Charge with Air as Dielectric

For a capacitor with air between its plates, the relationship between electric field (E), voltage (V), and plate separation (d) is given by: \(E = \frac{V}{d}\). Since E should not exceed \(3.00 \times 10^4\) V/m and \(d = 1.50\) mm, we calculate \(V\):\(V = E \times d = 3.00 \times 10^4\ \text{V/m} \times 1.50 \times 10^{-3}\ \text{m} = 45\ \text{V}\).The charge (Q) is calculated as: \(Q = C_0 \times V\), where \(C_0 = 8.00\ \text{pF} = 8.00 \times 10^{-12}\ \text{F}\).\(Q = 8.00 \times 10^{-12}\ \text{F} \times 45\ \text{V} = 3.60 \times 10^{-10}\ \text{C}\).

Calculate Max Charge with Dielectric

When a dielectric with a dielectric constant (K) is inserted, the capacitance becomes \(C = K \times C_0\). With \(K = 2.70\), the capacitance becomes:\(C = 2.70 \times 8.00\ \text{pF} = 21.6\ \text{pF} = 21.6 \times 10^{-12}\ \text{F}\).The electric field limit and plate separation remain unchanged. Therefore, the maximum allowable voltage across the plates is still 45 V. Thus, the new charge is:\(Q = C \times V = 21.6 \times 10^{-12}\ \text{F} \times 45\ \text{V} = 9.72 \times 10^{-10}\ \text{C}\).

Key concepts

Electric Field

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.

Dielectric Constant

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.

Parallel-plate Capacitor

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.

Charge Calculation

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.