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

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

Short Answer

Expert verified
(a) 5.65 × 10⁷ N/C; (b) 0.283 V, inside higher; (c) Use energy formula; (d) With dielectric: E'=1.046 × 10⁷ N/C, ΔV'=0.0523 V.

Step by step solution

01

Understanding Charge Density and E-Field Relationship

We need to calculate the electric field (\(E\)) in the wall between the two charged layers. First, we use Gauss's law, which relates electric charge and electric field. For an infinite plane with surface charge density \(\sigma\), the magnitude of the electric field just outside the surface is \(E = \dfrac{\sigma}{\varepsilon_0}\), where \(\varepsilon_0 = 8.85 \times 10^{-12}\) C\(^2\)/N m\(^2\) is the permittivity of free space.
02

Calculating the Electric Field in the Wall

Given charge density \(\sigma = 0.50 \times 10^{-3}\) C/m\(^2\), the magnitude of the electric field inside the wall is\[ E = \dfrac{0.50 \times 10^{-3}}{8.85 \times 10^{-12}} = 5.65 \times 10^{7} \text{ N/C}. \]
03

Finding the Potential Difference

The potential difference \(\Delta V\) is given by \(\Delta V = Ed\), where \(d\) is the thickness of the cell wall (5.0 nm or \(5.0 \times 10^{-9}\) m). Therefore,\[ \Delta V = 5.65 \times 10^7 \times 5.0 \times 10^{-9} = 0.283 \text{ V}. \] The inside surface has a higher potential because positive charges reside on the outside surface.
04

Estimating Total Electric-field Energy

The energy stored in an electric field per unit volume is given by \(u = \dfrac{1}{2}\varepsilon_0E^2\). For a spherical shell, the volume \(V\) of the wall is calculated using the outer and inner radius. Assume a single spherical cell, with radius \(r\) from volume \(\frac{4}{3}\pi r^3 = 10^{-16}\). The thickness is 5.0 nm, so the energy is \(U = u \times V\).
05

Adjusting for a Dielectric Layer

When the dielectric constant \(\kappa = 5.4\) is considered, the electric field inside the wall changes to \(E' = E/\kappa = 5.65 \times 10^7 / 5.4 = 1.046 \times 10^7 \text{ N/C}\). The potential difference is now\[ \Delta V' = E'd = 1.046 \times 10^7 \times 5.0 \times 10^{-9} = 0.0523 \text{ V}. \] The inside surface remains at a higher potential.

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.

Gauss's Law
Gauss's law is a fundamental principle in electrostatics that connects electric fields and electric charges. The law states that the total flux of an electric field through a closed surface is proportional to the enclosed electric charge. In mathematical terms, Gauss's law is expressed as: \[ \Phi = \oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enc}}{\varepsilon_0} \] Here, \( \Phi \) is the electric flux, \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area vector on the surface, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space. For practical calculations, Gauss's law is particularly useful in determining electric fields for symmetrical charge distributions, such as infinite planes, spheres, or cylinders. In this exercise, we apply Gauss's law to calculate the electric field between two parallel charge layers with known surface charge density.
Electric Field
The electric field (E) represents the force that a charge experiences in a region due to other electric charges. It is a vector quantity, which means it has both magnitude and direction. An important equation for electric fields with uniform surface charge density \( \sigma \) is given by: \[ E = \dfrac{\sigma}{\varepsilon_0} \] This equation indicates that the electric field is directly proportional to the charge density on the surface and inversely proportional to the permittivity of free space. In our exercise, this formula is used to find the magnitude of the electric field within the cell wall, exhibiting a parallel layer of negative and positive charges.
  • The electric field points from the positive to the negative charge.
  • Higher charge density results in a stronger electric field.
Understanding how electric fields operate between different charges helps explain how ions move across cell membranes in biological systems.
Potential Difference
The potential difference, often referred to as voltage (\(\Delta V\)), describes the work needed to move a unit charge from one point to another in an electric field. It can be calculated using the electric field and the distance between charges: \[ \Delta V = E \times d \] Where \(E\) is the electric field strength and \(d\) is the distance between the points, known as the thickness of the cell wall in this exercise. A higher potential indicates more energy per charge at a specific point, and in the context of the exercise, the inside of the cell is at a higher potential. This is essential in understanding how cells maintain ion gradients. Points to consider:
  • Potential difference determines the direction of charge flow.
  • A higher potential on the inner wall means charges will tend to move outwards.
Dielectric Constant
The dielectric constant, also known as relative permittivity (\(\kappa\)), measures a material's ability to reduce the electric field within it compared to the vacuum. It is defined as: \[ \kappa = \dfrac{\varepsilon}{\varepsilon_0} \] Where \( \varepsilon \) is the permittivity of the material and \( \varepsilon_0 \) is the permittivity of free space. Materials with a higher dielectric constant effectively reduce the electric field's strength, impacting how charge behaves within the material. In our exercise, when the cell wall is modeled with a dielectric constant of 5.4, the electric field inside is reduced, thereby decreasing the potential difference. This reflects the effect of the tissue on ion movement in biological cells:
  • High dielectric constant materials better insulate against electric fields.
  • Reduced electric field strength leads to lesser potential energy storage.

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

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm\(^2\). Each plate carries a charge of magnitude 4.35 \(\times\) 10\(^{-8}\) C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

A 20.0-\(\mu\)F capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-\(\mu\)F capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 \(\times\) 10\(^2\) V. The electric field between the plates is to be no greater than 1.00 \(\times\) 10\(^4\) N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 \(\mu\)C on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

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