Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

Short Answer

Expert verified
a) Charges on each capacitor: 7.61 × 10^{-10} C b) Total energy stored: 3.65 × 10^{-8} J c) Electric field between the plates of the capacitor with porcelain dielectric: 1.23 × 10^6 V/m

Step by step solution

01

Calculate the capacitances of \(C_1\) and \(C_2\)

We can calculate the capacitances using the formula \(C = \epsilon_0 \kappa \frac{A}{d}\), where \(\epsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{F/m}\)), \(\kappa\) is the dielectric constant (1 for air and 7 for porcelain), \(A\) is the area, and \(d\) is the separation between the capacitor plates. For \(C_1\), with air as dielectric and \(\kappa_{1} = 1\): \(C_1 = \epsilon_0 \kappa_{1} \frac{A}{d} = \frac{(8.85 \times 10^{-12} \mathrm{F/m})(1)(1 \times 10^{-4} \mathrm{m^2})}{1 \times 10^{-4} \mathrm{m}} = 8.85 \times 10^{-12} \mathrm{F}\) For \(C_2\), with porcelain as dielectric and \(\kappa_{2} = 7\): \(C_2 = \epsilon_0 \kappa_{2} \frac{A}{d} = \frac{(8.85 \times 10^{-12} \mathrm{F/m})(7)(1 \times 10^{-4} \mathrm{m^2})}{1 \times 10^{-4} \mathrm{m}} = 62.0 \times 10^{-12} \mathrm{F}\)
02

Calculate the equivalent capacitance of the series combination

The equivalent capacitance of two capacitors connected in series is given by: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) Plug in the values obtained in Step 1: \(\frac{1}{C_{eq}} = \frac{1}{8.85 \times 10^{-12}\, \text{F}} + \frac{1}{62.0 \times 10^{-12}\, \text{F}}\) Now, we can calculate the equivalent capacitance: \(C_{eq} = \frac{1}{\frac{1}{8.85 \times 10^{-12}\, \text{F}} + \frac{1}{62.0 \times 10^{-12}\, \text{F}}} = 7.93 \times 10^{-12} \, \text{F}\)
03

Calculate the charge on the equivalent capacitor

The charge \(Q_{eq}\) on the equivalent capacitor can be calculated using the formula \(Q_{eq} = C_{eq} \times V\), where \(V\) is the voltage supplied by the battery. \(Q_{eq} = (7.93 \times 10^{-12}\, \text{F})(96.0 \, \mathrm{V}) = 7.61 \times 10^{-10} \, \mathrm{C}\)
04

Determine the charges on each capacitor

As the capacitors are connected in series, the charge on each capacitor is the same as the charge on the equivalent capacitor. \(Q_{1} = Q_{2} = Q_{eq} = 7.61 \times 10^{-10} \, \mathrm{C}\)
05

Calculate the total energy stored in the capacitors

The total energy stored in the capacitors connected in series is given by: \(U_{total} = \frac{1}{2}C_{eq}\,V^2\) Plug in the values obtained for \(C_{eq}\) and \(V\): \(U_{total} = \frac{1}{2}(7.93 \times 10^{-12}\,\text{F})(96.0\,\mathrm{V})^2 = 3.65 \times 10^{-8} \, \mathrm{J}\)
06

Calculate the electric field between the plates of \(C_{2}\)

We can calculate the electric field \(E\) for \(C_2\) using the formula \(E = \frac{Q}{\epsilon_0 \kappa_2 A}\), where \(\epsilon_0\) is the permittivity of free space, \(\kappa_2\) is the dielectric constant of porcelain, and \(A\) is the area of the capacitor plates. \(E = \frac{7.61 \times 10^{-10} \mathrm{C}}{(8.85 \times 10^{-12} \mathrm{F/m})(7)(1 \times 10^{-4} \mathrm{m^2})} = 1.23 \times 10^6 \, \mathrm{V/m}\) The answers are: a) The charges on each capacitor are \(7.61 \times 10^{-10} \mathrm{C}\). b) The total energy stored in the capacitors is \(3.65 \times 10^{-8} \, \mathrm{J}\). c) The electric field between the plates of \(C_{2}\) is \(1.23 \times 10^6 \, \mathrm{V/m}\).

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Key Concepts

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

Capacitance Calculation
Understanding the capacitance of a parallel plate capacitor is foundational in studying electric fields and storing energy in an electrical system. Capacitance is the ability of a system to store an electric charge and is calculated by the formula

\( C = \epsilon_0 \kappa \frac{A}{d} \),

where \( \epsilon_0 \) represents the permittivity of free space, \( \kappa \) is the dielectric constant which quantifies the effect of the insulating material between the plates, \( A \) is the plate area, and \( d \) is the separation between the plates. In practice, to calculate it, simply plug in the values corresponding to the specific capacitor in question. A larger plate area or smaller separation distance results in higher capacitance, meaning the component can store more charge at a given voltage. For capacitors in series, the equivalent capacitance is found inversely by summing the reciprocals of individual capacitances and then taking the reciprocal of that sum.
Dielectric Constant
The dielectric constant, also known as the relative permittivity, signifies how much a dielectric material can reduce the electric field within a capacitor compared to the electric field in a vacuum. It is denoted by the Greek letter \( \kappa \). An interesting aspect of dielectrics is that they enhance a capacitor’s ability to store charge without affecting the size or separation of the plates. This happens because the dielectric reduces the electric field and, as a result, the potential difference for a given amount of charge. Consequently, it allows more charge to accumulate for the same voltage. Materials with a high dielectric constant, such as porcelain, significantly increase the capacitance when placed between the plates of a capacitor.
Energy in Capacitors
The stored electrical energy within capacitors is a critical concept in electric circuits. For a given capacitor, the energy (\( U \)) is given by the equation

\( U = \frac{1}{2} C V^2 \),

where \( C \) is the capacitance and \( V \) is the voltage across the capacitor. For series circuits, the energy is calculated using the equivalent capacitance of the system. Since energy is proportional to the square of the voltage, even a small increase in voltage leads to a much larger increase in stored energy. This highlights why managing voltage is crucial in circuit design to prevent energy waste and potential damage to electrical components.
Electric Field in Capacitors
The electric field within a parallel plate capacitor illustrates the intensity of an electric field as force per unit charge. It can be determined by the expression

\( E = \frac{Q}{\epsilon_0 \kappa A} \),

where \( Q \) is the charge stored on the capacitor, \( \epsilon_0 \) is the permittivity of free space, \( \kappa \) is the dielectric constant, and \( A \) represents the area of the plates. The electric field is inversely proportional to the dielectric constant; as \( \kappa \) increases, the electric field strength decreases. This is why using a material with a higher dielectric constant in the capacitor will lower the electric field strength for the same charge and area.

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Most popular questions from this chapter

Thermocoax is a type of coaxial cable used for high-frequency filtering in cryogenic quantum computing experiments. Its stainless steel shield has an inner diameter of \(0.35 \mathrm{~mm},\) and its Nichrome conductor has a diameter of \(0.17 \mathrm{~mm}\). Nichrome is used because its resistance doesn't change much in going from room temperature to near absolute zero. The insulating dielectric is magnesium oxide \((\mathrm{MgO}),\) which has a dielectric constant of \(9.7 .\) Calculate the capacitance per meter of Thermocoax.

Determine all the values of equivalent capacitance you can create using any combination of three identical capacitors with capacitance \(C\).

How much energy can be stored in a capacitor with two parallel plates, each with an area of \(64.0 \mathrm{~cm}^{2}\) and separated by a gap of \(1.30 \mathrm{~mm}\), filled with porcelain whose dielectric constant is \(7.0,\) and holding equal and opposite charges of magnitude \(420 . \mu C ?\)

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

Calculate the capacitance of the Earth. Treat the Earth as an isolated spherical conductor of radius \(6371 \mathrm{~km}\).

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