/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 The potential difference across ... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 114

The potential difference across a cell membrane from outside to inside is initially at \(-90 \mathrm{mV}\) (when in its resting phase). When a stimulus is applied, Na" ions are allowed to move into the cell such that the potential changes to \(+20 \mathrm{mV}\) for a short amount of time. (a) If the membrane capacitance per unit area is $1 \mu \mathrm{F} / \mathrm{cm}^{2},\( how much charge moves through a membrane of area \)0.05 \mathrm{cm}^{2} ?\( (b) The charge on \)\mathrm{Na}^{+}\( is \)+e$ How many ions move through the membrane?

Short Answer

Expert verified
How many sodium ions (Na+) move through the membrane, given that the charge on Na+ is +e? Answer: The amount of charge that moves through the membrane is 5.5 x 10^-9 C. Approximately 3.4 x 10^10 sodium ions move through the membrane.

Step by step solution

01

Calculate the change in potential difference

First, calculate the change in potential difference (voltage) by finding the difference between the final voltage (+20 mV) and the initial voltage (-90 mV): 螖V = V_final - V_initial 螖V = (20 - (-90)) mV 螖V = 110 mV
02

Convert potential difference to volts

Since we need to work in standard units (SI units), we need to convert the potential difference from millivolts (mV) to volts (V): 螖V = 110 mV * (1 V / 1000 mV) = 0.110 V
03

Calculate the charge moved through the membrane

Now, we can use the capacitance formula to find the charge (Q) that moves through the membrane: Q = C * 螖V Given the membrane capacitance per unit area (C/A) is 1 渭F/cm虏 and the area (A) is 0.05 cm虏, we can find the total capacitance (C) as follows: C = (C/A) * A C = (1 渭F/cm虏) * (0.05 cm虏) C = 0.05 渭F Now, convert the capacitance to standard units (farads) by multiplying by 1x10鈦烩伓 C = 0.05 渭F * (1 F / 10鈦� 渭F) = 5 x 10^-8 F Finally, calculate the charge (Q): Q = C * 螖V Q = (5 x 10^-8 F) * (0.110 V) Q = 5.5 x 10^-9 C Thus, the charge that moves through the membrane is 5.5 x 10^-9 C.
04

Calculate the number of sodium ions

Now that we have found the charge that moves through the membrane, we can find the number of sodium ions (N) by dividing the total charge (Q) by the charge of a single Na+ ion (+e = 1.6 x 10^-19 C): N = Q / e N = (5.5 x 10^-9 C) / (1.6 x 10^-19 C/ion) N 鈮� 3.4 x 10^10 ions Thus, approximately 3.4 x 10^10 sodium ions move through the membrane.

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Most popular questions from this chapter

A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$
It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?
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