/*! 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 107 The inside of a cell membrane is... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 107

The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?

Short Answer

Expert verified
Answer: The work done by the electric field on a sodium ion is -1.44 x 10^-20 J.

Step by step solution

01

Identify the given quantities

We are given the following information: - Potential difference between inside and outside of the cell membrane (\(\Delta V\)): \(-90.0 \mathrm{mV}\) - Charge of a sodium ion (\(q\)): \(+e\)
02

Convert the potential difference to volts (V)

The potential difference value is given in millivolts (mV). Convert it to volts (V) by dividing by 1000: \(\Delta V = -90.0 \mathrm{mV} \times \frac{1 \mathrm{V}}{1000 \mathrm{mV}} = -0.090 \mathrm{V}\)
03

Calculate the work done by the electric field

Use the formula for work done by an electric field, \(W=q\Delta V\), where \(q\) is the charge of the sodium ion and \(\Delta V\) is the potential difference: \(W = (+e)(-0.090 \mathrm{V})\) Here, \(e\) represents the elementary charge, which is approximately \(1.6 \times 10^{-19} \mathrm{C}\). Substitute this value into the equation: \(W = (1.6 \times 10^{-19}\mathrm{C})(-0.090 \mathrm{V})\)
04

Calculate the final answer

Multiply the values to get the work done by the electric field: \(W = -1.44 \times 10^{-20} \mathrm{J}\) The electric field does \(-1.44 \times 10^{-20} \mathrm{J}\) of work on the sodium ion as it moves through the cell membrane from outside to inside. The negative sign indicates that the work is done against the electric force, which means that the work is done on the sodium ion.

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Most popular questions from this chapter

Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
An axon has the outer part of its membrane positively charged and the inner part negatively charged. The membrane has a thickness of \(4.4 \mathrm{nm}\) and a dielectric constant \(\kappa=5 .\) If we model the axon as a parallel plate capacitor whose area is \(5 \mu \mathrm{m}^{2},\) what is its capacitance?
A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?
A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
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