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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 ± 0.50 × 103 C/m2, the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of 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 1016 m3. 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 σ, the magnitude of the electric field just outside the surface is E=σε0, where ε0=8.85×1012 C2/N m2 is the permittivity of free space.
02

Calculating the Electric Field in the Wall

Given charge density σ=0.50×103 C/m2, the magnitude of the electric field inside the wall isE=0.50×1038.85×1012=5.65×107 N/C.
03

Finding the Potential Difference

The potential difference ΔV is given by ΔV=Ed, where d is the thickness of the cell wall (5.0 nm or 5.0×109 m). Therefore,ΔV=5.65×107×5.0×109=0.283 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=12ε0E2. 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 43πr3=1016. The thickness is 5.0 nm, so the energy is U=u×V.
05

Adjusting for a Dielectric Layer

When the dielectric constant κ=5.4 is considered, the electric field inside the wall changes to E=E/κ=5.65×107/5.4=1.046×107 N/C. The potential difference is nowΔV=Ed=1.046×107×5.0×109=0.0523 V. The inside surface remains at a higher potential.

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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: Φ=EdA=Qencε0 Here, Φ is the electric flux, E is the electric field, dA is the differential area vector on the surface, Qenc is the enclosed charge, and ε0 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 σ is given by: E=σε0 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 (ΔV), 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: ΔV=E×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 (κ), measures a material's ability to reduce the electric field within it compared to the vacuum. It is defined as: κ=εε0 Where ε is the permittivity of the material and ε0 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.

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Most popular questions from this chapter

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

Your electronics company has several identical capacitors with capacitance C1 and several others with capacitance C2. You must determine the values of C1 and C2 but don't have access to C1 and C2 individually. Instead, you have a network with C1 and C2 connected in series and a network with C1 and C2 connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that C1 is greater than C2. (a) Calculate C1 and C2. (b) For the series combination, does C1 or C2 store more charge, or are the values equal? Does C1 or C2 store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm2. Each plate carries a charge of magnitude 4.35 × 108 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 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, what is the charge per unitlength λ on the capacitor?

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