Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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?

Short Answer

Expert verified
(a) Capacitance is approximately 1.18 µF/cm². (b) Electric field is about 1.13 × 10⁷ V/m.

Step by step solution

01

Understanding the problem

We need to find the capacitance per square centimeter of a cell membrane modeled as a parallel-plate capacitor and then determine the electric field inside it at a given potential difference.
02

Identify the formula for capacitance

The formula for capacitance of a parallel-plate capacitor is \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric material, \( A \) is the area, and \( d \) is the distance between the plates.
03

Calculate the permittivity

The relative permittivity (dielectric constant) \( \kappa \) is given as 10. The permittivity of the dielectric material is \( \varepsilon = \kappa \varepsilon_0 \),where \( \varepsilon_0 \approx 8.85 \times 10^{-12} \ \text{F/m} \) is the permittivity of free space.
04

Substitute values for capacitance

Given that \( \kappa = 10 \), \( \varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \), \( d = 7.5 \ \text{nm} = 7.5 \times 10^{-9} \ \text{m} \), substitute these values into the capacitance formula for unit area:\[ C = \frac{10 \times 8.85 \times 10^{-12} \ \text{F/m} \times 1 \ \text{cm}^2}{7.5 \times 10^{-9} \ \text{m}}. \]
05

Conversion of units

Convert \( 1 \ \text{cm}^2 \) to square meters: \( 1 \ \text{cm}^2 = 10^{-4} \ \text{m}^2 \). Substitute into the equation:\[ C = \frac{10 \times 8.85 \times 10^{-12} \times 10^{-4}}{7.5 \times 10^{-9}}. \]
06

Calculate the capacitance

Perform the calculations:\[ C \approx \frac{8.85 \times 10^{-15}}{7.5 \times 10^{-9}} = 1.18 \times 10^{-6} \ \text{F/cm}^2 = 1.18 \ \text{µF/cm}^2. \]
07

Identify the formula for electric field

The electric field \( E \) inside the membrane is given by \( E = \frac{V}{d} \) where \( V \) is the potential difference, and \( d \) is the thickness of the membrane.
08

Substitute values for electric field

Given \( V = 85 \ \text{mV} = 85 \times 10^{-3} \ \text{V} \) and \( d = 7.5 \times 10^{-9} \ \text{m} \), substitute into the formula:\[ E = \frac{85 \times 10^{-3}}{7.5 \times 10^{-9}}. \]
09

Calculate the electric field

Perform the calculation to find the electric field:\[ E \approx \frac{85 \times 10^{-3}}{7.5 \times 10^{-9}} \approx 1.13 \times 10^{7} \ \text{V/m}. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Parallel-Plate Capacitor
A cell membrane can be thought of as a parallel-plate capacitor. This model helps us understand how the membrane can store and separate charges. In a parallel-plate capacitor, two large plates face each other and are separated by a small distance.
  • The plates hold equal and opposite charges.
  • The distance between them is small compared to their size.
This arrangement allows for effective charging and energy storage. In the case of a cell membrane, the inner and outer surfaces of the membrane act as the plates. This configuration plays a crucial role in the membrane's ability to function in various biological processes, such as nerve impulses.
Dielectric Constant
The dielectric constant is an important factor when examining capacitors, including our cell membrane model. A dielectric is a non-conductive material placed between the plates of a capacitor. It increases the capacitance by reducing the electric field.
  • It's represented by \( \kappa \) (kappa).
  • A higher dielectric constant means a greater ability to store charge.
The cell membrane has a dielectric constant of about 10. This is due to its composition of proteins and organic material. The higher dielectric constant in the membrane enhances its capacity to store electrical energy, which is essential for its function as a biological capacitor.
Electric Field
Inside a cell membrane modeled as a capacitor, there exists an electric field. This field is crucial as it prevents further charges from crossing the membrane.
  • The electric field (E) is calculated using the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference, and \( d \) is the distance between the plates (thickness of the membrane).
  • An increased electric field makes it more challenging for ions to pass through, controlling the movement of charged particles.
For the tested membrane, the electric field is approximately \( 1.13 \times 10^{7} \ \text{V/m} \). This strong electric field is a vital part of cellular functions like signalling and nutrient transport.
Potential Difference
In a cell membrane, the potential difference is the difference in electrical potential between its inner and outer surfaces. In our model, it directly affects the electric field strength across the membrane.
  • This difference in potential is crucial for many biological processes, such as action potentials in neurons.
  • For the cell membrane example, the potential difference is \( 85 \ \text{mV} \).
The potential difference indicates how much energy per charge is available to move across the membrane. It acts as a driving force that pushes ions from one side of the membrane to another, enabling essential cellular activities.
Permittivity of Free Space
Permittivity of free space is a fundamental constant in physics, denoted by \( \varepsilon_0 \). It's the measure of resistance encountered when forming an electric field in a vacuum.
  • The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \ \text{F/m} \).
  • In capacitors, it's used in calculating capacitance, such as \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is permittivity, \( A \) is area, and \( d \) is distance.
For our cell membrane, the permittivity of free space is multiplied by the dielectric constant to find the effective permittivity. This results in the ability of the membrane to store and separate charges effectively, an essential attribute for its biological functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free