Question

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

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

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 \(\kappa_{2} = 7\): \(C_2 = \epsilon_0 \kappa_{2} \frac{A}{d} = \frac{(8.85 \times 10^{-12} \mathrm{F/m})(7)(1 \times 10^{-4} \mathrm{m^2})}{1 \times 10^{-4} \mathrm{m}} = 62.0 \times 10^{-12} \mathrm{F}\)

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: \(C_{eq} = \frac{1}{\frac{1}{8.85 \times 10^{-12}\, \text{F}} + \frac{1}{62.0 \times 10^{-12}\, \text{F}}} = 7.93 \times 10^{-12} \, \text{F}\)

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. \(Q_{eq} = (7.93 \times 10^{-12}\, \text{F})(96.0 \, \mathrm{V}) = 7.61 \times 10^{-10} \, \mathrm{C}\)

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. \(Q_{1} = Q_{2} = Q_{eq} = 7.61 \times 10^{-10} \, \mathrm{C}\)

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 \(V\): \(U_{total} = \frac{1}{2}(7.93 \times 10^{-12}\,\text{F})(96.0\,\mathrm{V})^2 = 3.65 \times 10^{-8} \, \mathrm{J}\)

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 \(1.23 \times 10^6 \, \mathrm{V/m}\).

Key concepts

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.