Problem 37 - Physics

Question

The membrane that surrounds a certain type of living cell has a surface area 8 of \(5.0 \times 10^{-9} \mathrm{~m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{~m}\). Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to \(\mathrm{K}^{+}\) ions (charge \(+e\) ), how many such ions are present on the outer surface?

Step-by-Step Solution

Answer

Charge on outer surface: \(1.326 \times 10^{-14} \ \text{C}\); \(\mathrm{K}^{+}\) ions: \(8.29 \times 10^4\) ions.

Step 1: Understand the Parallel Plate Capacitor Model

The membrane acts like a parallel plate capacitor, where the surface area of the plates is given as \(A = 5.0 \times 10^{-9} \ \text{m}^2\) and the distance between the plates (thickness) is \(d = 1.0 \times 10^{-8} \ \text{m}\). The dielectric constant \(k\) is \(5.0\).

Step 2: Use the Capacitor Formula to Find Capacitance

For a parallel plate capacitor, the capacitance \(C\) can be calculated using the formula: \[ C = \frac{\varepsilon_0 \cdot k \cdot A}{d} \] where \(\varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m}\) is the permittivity of free space.

Step 3: Substitute Values to Calculate Capacitance

Substitute the known values into the capacitance formula: \[ C = \frac{(8.85 \times 10^{-12}) \cdot 5.0 \cdot (5.0 \times 10^{-9})}{1.0 \times 10^{-8}} \] Compute this expression to find \(C\).

Step 4: Calculate the Capacitance

After computation, the capacitance \(C\) is: \[ C = 2.21 \times 10^{-13} \ \text{F} \]

Step 5: Use Capacitance to Find Charge

The charge \(Q\) on the outer surface can be found using the relationship between charge, capacitance, and voltage: \[ Q = C \cdot V \] where \(V = 60 \ \text{mV} = 60 \times 10^{-3} \ \text{V}\).

Step 6: Substitute Values to Calculate Charge

Substitute the values into the charge formula: \[ Q = (2.21 \times 10^{-13}) \cdot (60 \times 10^{-3}) \] Compute this expression to find \(Q\).

Step 7: Calculate Charge on Outer Surface

After computation, the charge \(Q\) on the outer surface is: \[ Q = 1.326 \times 10^{-14} \ \text{C} \]

Step 8: Determine Number of Ions

To find the number of \(\mathrm{K}^{+}\) ions, use the formula: \[ = \frac{Q}{e} \] where \(e = 1.6 \times 10^{-19} \ \text{C}\) is the charge of a single electron (and thus the charge of \(\mathrm{K}^{+}\)).

Step 9: Calculate Number of Ions

Substitute the charge \(Q\) and electron charge \(e\) into the formula: \[ n = \frac{1.326 \times 10^{-14}}{1.6 \times 10^{-19}} \] Compute this expression to find the number \(n\) of \(\mathrm{K}^{+}\) ions.

Step 10: Calculate the Resulting Number of Ions

After computation, the number of \(\mathrm{K}^{+}\) ions on the outer surface is approximately \(8.29 \times 10^4\) ions.

Key Concepts

Dielectric Constant

The dielectric constant is a measure of a material's ability to store electrical energy in an electric field. It is also known as the relative permittivity and is denoted by the symbol \( k \). In a parallel plate capacitor, the dielectric constant is crucial as it enhances the capacitor’s capacitance by reducing the electric field strength needed to store the same amount of electric charge.

The higher the dielectric constant, the more charge the capacitor can store. In the context of a cell membrane, which acts like a capacitor, a dielectric constant of 5.0 significantly increases its ability to store ions when compared to a vacuum, whose dielectric constant is 1. This plays a vital role in biological functions, as cell membranes rely on potentials created by stored charge to transmit signals.

Cell Membrane

A cell membrane is a biological structure that separates the interior of all cells from the outside environment. It consists mainly of lipids and proteins and acts not only as a barrier but also as an active and selective gateway for ions and molecules that enter or leave the cell.

In physics, a cell membrane can be compared to a parallel plate capacitor in the way it manages electric charges. This means that it has two conductive surfaces separated by an insulating layer, which is the lipid bilayer of the membrane. This separation allows the membrane to control voltages across it, essential for processes like nerve impulse transmission. The thickness and surface area of cell membranes are key factors influencing their capacitance and, consequently, their ability to store electric charge.

Electric Charge

Electric charge refers to the property of matter that causes it to experience a force when placed in an electric or magnetic field. There are two types of charges: positive and negative. When it comes to capacitors, including those mimicking cell membranes, they store electrical energy by holding different charges on each of their "plates."

• A thin cell membrane has the ability to separate charges, creating electrical potentials.
• The greater the potential difference, the more significant the charge separation, resulting in higher stored electric energy.

In the context of a cell membrane acting as a capacitor, charges such as potassium ions (\( \mathrm{K}^+ \)) accumulate on the surface. This charge accumulation is essential for physiological processes like nerve signaling and muscle contraction.

Permittivity of Free Space

The permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental physical constant characteristic of the vacuum. Its value is approximately \( 8.85 \times 10^{-12} \ ext{F/m} \) (farads per meter). This constant plays a crucial role in electromagnetism and capacitor design.

• It represents how much electric field is permitted in a vacuum.
• It's used in calculating the capacitance of capacitors, including those that emulate biological cell membranes.

In the calculation of capacitance for the cell membrane in the aforementioned problem, \( \varepsilon_0 \) is a fundamental part of the equation: \[ C = \frac{\varepsilon_0 \cdot k \cdot A}{d} \]. Understanding \( \varepsilon_0 \) helps in grasping how capacitors work and the amount of charge they can store.