coulombs-law-states-that-the-force-of-attraction-between-two-charged-particles-is-directly-proportio

Question

Coulomb's Law states that the force of attraction between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The figure shows particles with charge $ 1 $ located at positions $ 0 $ and $ 2 $ on the coordinate line and a particle with charge $ -1 $ at a position $ x $ between them. It follows from Coulomb's Law that the net force acting on the middle particle is
$$ F(x) = -\dfrac{k}{x^2} + \dfrac{k}{(x - 2)^2} $$ $$  0 < x < 2 $$
where $ k $ is a positive constant. Sketch the graph of the net force function. What does the graph say about the force?

Jamie M · Accepted Answer

So this is our net force function, and we have a domain from 0 to 2. All right, so now we can look it of prime. Negative K over X squared. Plus. Okay, over X minus two squared. Okay, this is inconclusive. If we try to set a equal to zero and find critical points, we see that it is inconclusive. Well, we can look att f double prime for Akane cavity in case I left a will. Prime to kay over X cubed minus two K over X minus two. Cute. And we see that this is monotone tickly increasing over our dummy. All right. And we also have vertical ass and tools. X equals zero and X equals two. Here. We can tell. Um, X equals zero in here. We can tell X equals two. When we said the denominators to zero. Okay. And I&#x27;m a Ken Griff over are doing all right, so we know that we do have an intercept 01 So let&#x27;s say this is one and that&#x27;s to this 01 There isn&#x27;t intercept right here. And we know that we have at Assen Toto X equals zero and X equals two. And We know that, um, this is not a Miramax, because this was inconclusive. So the first derivative with inconclusive and it&#x27;s monotone tickly increasing. Therefore, it has to look sure that little better, therefore has to look like this monotone tickly increasing and close to but not touching the ass And toots. So now let&#x27;s talk about the middle particle. So X equals one on the net. Force zero. Um, but if it&#x27;s greater than one is gonna be a positive net force, and the closer it gets to this, ask himto the faster is going to increase. And here, um, as X is less than one, the net force is gonna be negative. And the courtroom, just to the ask himto the faster it&#x27;s going to get decrease.

Problem

Find an equation of the slant asymptote. Do not s…

Problem 60 Easy Difficulty

Answer

When the middle particle is at $x=1,$ the net force acting on it is $0 .$ When
$x>1,$ the net force is positve, increasing more and more rapidly as the
particle approaches $x=2 .$ The opposite happens when $x<1$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Heather Z.

Kayleah T.

Caleb E.

Samuel H.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Calculus: Early Transcendentals

Heather Z.

Kayleah T.

Caleb E.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

When the middle particle is at $x=1,$ the net force acting on it is $0 .$ When$x>1,$ the net force is positve, increasing more and more rapidly as theparticle approaches $x=2 .$ The opposite happens when $x<1$

Calculus: Early Transcendentals

Heather Z.

Kayleah T.

Caleb E.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Heather Z.

Kayleah T.

Caleb E.

Samuel H.

When the middle particle is at $x=1,$ the net force acting on it is $0 .$ When
$x>1,$ the net force is positve, increasing more and more rapidly as the
particle approaches $x=2 .$ The opposite happens when $x<1$