The membrane that surrounds a certain type of living cell...

The membrane that surrounds a certain type of living cell has a surface area 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?

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor is a system of two conductive plates separated by a dielectric material, where the geometry (area and separation) directly influences its ability to store electrical energy. The capacitance, being proportional to the plate area and inversely proportional to the distance between them, is a fundamental concept in understanding how electric fields interact with materials in capacitive systems.

Capacitance

Capacitance is the measure of a capacitor's ability to store charge per unit voltage, defined by the formula C = ?? ?r A/d, where ?? is the vacuum permittivity, ?r is the relative dielectric constant, A is the area of the plates, and d is the separation between the plates. This relationship shows how material properties and physical dimensions affect the performance of capacitive devices.

Relationship Between Charge and Potential Difference

The stored charge on a capacitor is directly related to its capacitance and the potential difference across it, as given by the equation Q = C·V. This principle allows one to determine the amount of electric charge stored when a voltage is applied, making it a crucial concept in both theoretical and practical electrostatics.

Ionic Charge Quantization

Ionic charge quantization refers to the concept that electrical charge is carried by discrete particles, such as ions, each having a fixed elementary charge. By dividing the total charge by the charge of one ion, one can determine the number of ions involved, which highlights the quantized nature of charge in electrochemical and biophysical systems.

Transcript

00:01 Hi there, so for this problem we have a membrane that has a surface area a equals to 5 times 10 to the minus 9 meters squared and the thickness of this membrane, this thickness in here we're not going to call that thickness is equal to 1 times 10 to the minus 8 meters.

00:38 Now we need to assume that this membrane behaves like a part of the plate capacitor.

00:47 And then it has a dielectric constant of phi.

00:56 And for the first part of this problem, we are given the information that the potential on the other surface it is given and that is 60 millipoles and what we need to calculate is how much charge resides on the outer surface.

01:27 Now to do this, we know that the charge that resides on the other surface of the cell membrane is equal to the capacitance times the voltage.

01:47 Now, before we can use this expression to calculate that charge in here, we must further determine the capacitance of this membrane because that is not a given value in here.

02:04 So we note that definition of a capacitance is equal to the dilectric constants times the epsilon sub -zero, the vacuum permissivity, times the area of each plate over the distance that separate them.

02:36 Now, we have all of these values because d corresponds to the thickness of this membrane.

02:46 A, we can obtain, we also have the area, the surface area, and the other two are as constants, the dielectric constant is also given.

03:02 So we need to substitute all those values in here so to find the capacitance.

03:07 So we plot in here, the capacitance, the electric constant that is five times the vacuum permissivity, which is 8 .85 times 10 to the minus 12 ferrets per meter.

03:27 This times the area, the surface area that is also given for this problem, and it is 5 times 10 to the minus 9 meters squared, and all of this over the thickness or the distance between the plates that in this case correspond to the thickness, and that is equal to 1 times 10 to the minus 8 meters.

03:53 So plotting all of this into the calculator, we obtain that the capacitance for this membrane is 2 .2 times 10 to the minus 11 forrests...

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