How Does Changing Distance Affect Force Between Charges?

In summary, to change the force between two point charges by a factor of ten, you must change the distance between them by a factor of 1/√10.
  • #1
dummie90
1
0

Homework Statement


By what factor must you change the distance between two point charges to change the force between them by a factor of ten?


Homework Equations


F1=kq1/r^2
F2=kq2/r^2



The Attempt at a Solution


F1:F2= 10
r1/r2=sq.root(F2/F1)= 1/sqroot10

I do not understand how to get to r1/r2=sq.root(F2/F1)= 1/sqroot10
I do understand that you square root F2/F1 to get rid of the square of r. but why is F2 on top of the fraction instead of F1 (i.e. sqroot F1/F2) and how does that become 1/sqroot 10? And how did that equation even come about?

Sorry I'm very confused I've tried to make sense of this worked example many times,
Thank you so much for the help
 
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  • #2
Welcome to PF!

dummie90 said:
F1=kq1/r^2
F2=kq2/r^2

Sorry I'm very confused …

Hi dummie90! Welcome to PF! :smile:

(have a square-root: √ :smile:)

You're very confused because your equations are wrong. :wink:

They should be:

F1=kqQ/r1^2
F2=kqQ/r2^2

now can you see how it comes out as √(F2/F1)? :smile:
 
  • #3
.

Hello, thank you for reaching out for clarification. Let's break down the equation step by step to better understand it.

First, we have the equation F1=F2, which means that the force between the two charges must be equal in both scenarios (before and after changing the distance).

Next, we have the equation F=kq1q2/r^2, which is Coulomb's Law. This equation tells us that the force between two charges is directly proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance between them (r^2).

Now, in order to change the force by a factor of ten, we need to change the distance by a certain factor. Let's call this factor x. This means that the new distance will be x times the original distance (r2=x*r1).

Substituting this into Coulomb's Law, we get F2=kq1q2/(x*r1)^2. Simplifying, we get F2=kq1q2/(x^2*r1^2).

Now, we can see that F2 is on top of the fraction because we are looking for the new force (F2) after changing the distance. We want to find the value of x that will make F2 ten times larger than F1. In other words, we want to find the value of x that will make F2=10F1. So we can rewrite the equation as 10F1=kq1q2/(x^2*r1^2).

Next, we can divide both sides by F1 to get 10=kq1q2/(F1*x^2*r1^2). Since we know that F1=kq1q2/r1^2, we can substitute this in to get 10=(F1*x^2*kq1q2)/(F1*x^2*r1^2). Simplifying, we get 10=x^2, which means that x=sqrt(10).

Therefore, to change the force between two point charges by a factor of ten, we need to change the distance between them by a factor of sqrt(10), or approximately 3.16. I hope this helps clarify the equation for you.
 

What is electricity?

Electricity is a physical phenomenon that involves the flow of electric charge. It is responsible for the functioning of many devices and systems, including lighting, electronics, and power generation.

What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two charged particles. It states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

What is the unit of electric charge?

The unit of electric charge is the coulomb (C). It is defined as the amount of charge that passes through a conductor in one second when a current of one ampere is flowing.

How does Coulomb's Law relate to electric fields?

Coulomb's Law is related to electric fields as it explains the force between two charges that create or interact with the electric field. The magnitude and direction of the electric field can be calculated using Coulomb's Law and the values of the charges involved.

What are some real-life applications of Coulomb's Law?

Coulomb's Law has many real-life applications, such as in the functioning of electronic devices, power generation and distribution, and even in medical technologies such as MRI machines. It is also used in the study of particle physics and in understanding the behavior of atoms and molecules.

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