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Chapter 20: Problem 65

Cell Membranes The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area \(4.75 \times 10^{-9} \mathrm{m}^{2},\) a plate separation of \(8.5 \times 10^{-9} \mathrm{m},\) and a dielectric with a dielectric constant of \(4.5 .\) (a) What is the energy stored in such a cell membrane if the potential difference across it is \(0.0725 \mathrm{V}\) ? (b) Would your answer to part (a) increase, decrease, or stay the same if the thickness of the cell membrane is increased? Explain.

Short Answer

Expert verified
(a) Energy is calculated using the capacitance formula and potential difference; (b) the stored energy decreases if membrane thickness increases.

Step by step solution

01

Calculate the Capacitance

To determine the energy stored in the capacitor, we must first find its capacitance. The formula for the capacitance of a parallel-plate capacitor is given by: \[ C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \) (the permittivity of free space), \( \varepsilon_r = 4.5 \) (the dielectric constant), \( A = 4.75 \times 10^{-9} \ \mathrm{m}^2 \) (the area of the plates), and \( d = 8.5 \times 10^{-9} \ \mathrm{m} \) (the separation between the plates). Substitute these values into the formula to calculate \( C \).
02

Compute the Capacitance Value

Plug in the given values into the formula for capacitance: \[ C = (8.85 \times 10^{-12}) \cdot 4.5 \cdot \frac{4.75 \times 10^{-9}}{8.5 \times 10^{-9}} \] Calculate the capacitance \( C \).
03

Calculate the Energy Stored

Once the capacitance \( C \) is determined, use the formula for the energy stored in a capacitor: \[ U = \frac{1}{2} C V^2 \] where \( V = 0.0725 \ \mathrm{V} \). Substitute the value of \( C \) from Step 2 and the given \( V \) to find the energy stored, \( U \).
04

Discuss the Effect of Plate Separation on Energy

According to the formula for capacitance \( C \), increasing the plate separation \( d \) while keeping other factors constant will decrease \( C \), as \( C \) is inversely proportional to \( d \). Since the stored energy \( U = \frac{1}{2} C V^2 \), a decrease in \( C \) leads to a decrease in \( U \). Therefore, if the thickness (plate separation) is increased, the energy stored will decrease.

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a simple type of capacitor used to store electrical energy. It consists of two conductive plates separated by a certain distance. When a voltage is applied across these plates, they allow for the storage of charge. The formula to calculate the capacitance \( C \) of a parallel-plate capacitor is \( C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \). Here, \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) is the dielectric constant, \( A \) is the plate area, and \( d \) is the separation between the plates.
  • Area \( A \): Larger areas can store more charge, increasing the capacitance.
  • Distance \( d \): Greater separation reduces capacitance.
Capacitors are widely used in electronic devices and have numerous applications in the storage and management of electrical energy.
Dielectric Constant
The dielectric constant, denoted as \( \varepsilon_r \), is a measure of a material's ability to increase the capacitance of a capacitor compared to a vacuum. It represents how much the electric field within a material is reduced compared to the field in a vacuum. For example, in the given exercise, the dielectric constant is 4.5.
  • Effect on Capacitance: Higher dielectric constants result in greater capacitance, since they allow for more charge to be stored at a given voltage.
  • Material Role: The material between the plates of the capacitor can significantly impact performance by increasing the overall capacitance.
Choosing the right dielectric material enhances the effectiveness of capacitors in various applications.
Stored Energy
Stored energy in a capacitor is the energy accumulated due to charges separated by an electric potential. It's calculated using the formula \( U = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage. In practical applications, the capacity to store and release energy quickly makes capacitors ideal for many electronic tasks.
  • Dependence on Capacitance: Energy storage increases with higher capacitance since more charge can be stored.
  • Voltage Influence: The potential difference \( V \) significantly affects stored energy, as seen in the quadratic relationship \( V^2 \).
Understanding how energy is stored in capacitors is crucial for designing circuits with the desired performance characteristics.
Electric Potential
Electric potential, often referred to as voltage, is the potential energy per charge unit. In a capacitor, this potential difference is what allows charges to do work. It is critical for determining how much energy is stored, as illustrated by the formula \( U = \frac{1}{2} C V^2 \).
  • Potential Difference \( V \): Defines the amount of energy transformed or transferred per charge moved between two points.
  • Impact on Operations: Variation in the potential difference leads to changes in stored energy in capacitors, affecting their performance.
Understanding electric potential is essential to harness the full capabilities of capacitors in electrical circuits.

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Most popular questions from this chapter

Two point charges are on the \(x\) axis. Charge 1 is \(+q\) and is located at \(x=-1.0 \mathrm{m} ;\) charge 2 is \(+2 q\) and is located at \(x=1.0 \mathrm{m}\). Make sketches of the equipotential surfaces for this system (a) out to a a distance of about \(2.0 \mathrm{m}\) from the origin and (b) far from the origin. In each case, indicate the direction in which the potential increases.

A proton is released from rest in a region of space with a nonzero electric field. As the proton moves, does it experience an increasing or decreasing electric potential? Explain.

A charge of \(-2.205 \mu C\) is located at \((3.055 \mathrm{m}, 4.501 \mathrm{m}),\) and a charge of \(1.800 \mu C\) is located at \((-2.533 \mathrm{m}, 0)\) (a) Find the electric potential at the origin. (b) There is one point on the line connecting these two charges where the potential is zero. Find this point.

Cell Membranes and Dielectrics Many cells in the body have a cell membrane whose inner and outer surfaces carry opposite charges, just like the plates of a parallel-plate capacitor. Suppose a typical cell membrane has a thickness of \(8.1 \times 10^{-9} \mathrm{m},\) and its inner and outer surfaces carry charge densities of \(-0.58 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2}\) and \(+0.58 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) respectively. In addition, assume that the material in the cell membrane has a dielectric constant of \(5.5 .\) (a) Find the direction and magnitude of the electric field within the cell membrane. (b) Calculate the potential difference between the inner and outer walls of the membrane, and indicate which wall of the membrane has the higher potential.

A parallel-plate capacitor filled with air has plates of area \(0.0066 \mathrm{m}^{2}\) and a separation of \(0.45 \mathrm{mm} .\) (a) Find the magnitude of the charge on each plate when the capacitor is connected to a 12-V battery. (b) Will your answer to part (a) increase, decrease, or stay the same if the separation between the plates is increased? Explain. (c) Calculate the magnitude of the charge on the plates if the separation is \(0.90 \mathrm{mm}\).
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