Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 capacitance of \(1.20 \mathrm{nF}\) There is a charge of magnitude \(0.800 \mu\) C on each plate. (a) What is the potential difference between the plates? (b) If the plate separation is doubled, while the charge is kept constant, what will happen to the potential difference?
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?
Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
The bottom of a thundercloud is at a potential of $-1.00 \times 10^{8} \mathrm{V}\( with respect to Earth's surface. If a charge of \)-25.0 \mathrm{C}$ is transferred to the Earth during a lightning strike, find the electric potential energy released. (Assume that the system acts like a capacitor-as charge flows, the potential difference decreases to zero.)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Scientific Method Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free