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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

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

Step by step solution

01

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.
02

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 \(C_0 = 8.00\ \text{pF} = 8.00 \times 10^{-12}\ \text{F}\).\(Q = 8.00 \times 10^{-12}\ \text{F} \times 45\ \text{V} = 3.60 \times 10^{-10}\ \text{C}\).
03

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:\(Q = C \times V = 21.6 \times 10^{-12}\ \text{F} \times 45\ \text{V} = 9.72 \times 10^{-10}\ \text{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

A cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

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