/*! 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 17 \(\mathrm{A} \mathrm{K}^{+}\) io... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 17

\(\mathrm{A} \mathrm{K}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion are directly across from each other on opposite sides of a membrane 9.0 nm thick. What is the electric force on the \(\mathrm{K}^{+}\) ion due to the \(\mathrm{Cl}^{-}\) ion? Ignore the presence of other charges.

Short Answer

Expert verified
Answer: The electric force on the K\(^{+}\) ion due to the Cl\(^{-}\) ion is approximately \(4.3 \times 10^{-12}\,\mathrm{N}\).

Step by step solution

01

The given ions are a \(\mathrm{K}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion. The charge of each ion: \(q_1 = +e\) (for K\(^{+}\)) \(q_2 = -e\) (for Cl\(^{-}\)) Where \(e = 1.602 \times 10^{-19}\,\mathrm{C}\) (elementary charge) #step2# - Convert the distance between the ions into meters

The distance between the ions is given as 9.0 nm. We need to convert it into meters: \(r = 9.0 \,\mathrm{nm} = 9.0 \times 10^{-9}\, \mathrm{m}\) #step3# - Calculate the force using Coulomb's law
02

Now, we plug the charges and distance into the Coulomb's law formula to find the electric force: \(F = k\frac{(+e)(-e)}{(9.0 \times 10^{-9}\,\mathrm{m})^2}\) \(F = \frac{(8.9875脳10^9\,\mathrm{Nm^2/C^2})(1.602 \times 10^{-19}\,\mathrm{C})^2}{(9.0 \times 10^{-9}\,\mathrm{m})^2}\) #step4# - Compute the force

By calculating the above expression, we find the electric force: \(F \approx 4.3 \times 10^{-12}\, \mathrm{N}\) Hence, the electric force on the \(\mathrm{K}^{+}\) ion due to the \(\mathrm{Cl}^{-}\) ion is approximately \(4.3 \times 10^{-12}\,\mathrm{N}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Find the total positive charge of all the protons in \(1.0 \mathrm{mol}\) of water.
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
Suppose a 1.0 -g nugget of pure gold has zero net charge. What would be its net charge after it has \(1.0 \%\) of its electrons removed?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Turning Points in Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Measurements

Read Explanation

Electrostatics

Read Explanation

Waves Physics

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.