F = kQq/R2 finding force problem

  • Thread starter xNick94
  • Start date
  • Tags
    Force
In summary: This is a difficult problem to perform the algebraic steps to get to the answer, but the most basic concept to understand is that the force between two charges is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force decreases. The equation F = kQq/R^2 represents this relationship, where k is a constant, Q and q are the charges of the two objects, and R is the distance between them.In this problem, we are given the initial force (3.1x10^-
  • #1
xNick94
8
0
Two charges Q1 amd Q2 jabe a force of 3.1x10^-8 N when the distance between them is 3m. Use the formula F = kQq/R2 to determine the force when the distance (R) between charge Q1 and Q2 increased from 3m to 4.5m

The Attempt at a Solution



My attempt was that
F = 1/R^2
k= 9.0x10^9

F=(3/4.5)^2
F=0.44N

3.1x10^-8 / 0.44
=.7 x10^-8

The answer being F = .7 x10^-8 N


Is that correct?
 
Physics news on Phys.org
  • #2
xNick94 said:
Two charges Q1 amd Q2 jabe a force of 3.1x10^-8 N when the distance between them is 3m. Use the formula F = kQq/R2 to determine the force when the distance (R) between charge Q1 and Q2 increased from 3m to 4.5m

The Attempt at a Solution



My attempt was that
F = 1/R^2
k= 9.0x10^9

F=(3/4.5)^2
F=0.44N

3.1x10^-8 / 0.44
=.7 x10^-8

The answer being F = .7 x10^-8 N


Is that correct?

Up to (3/4.5)^2 I was fairly comfortable with.

From there on, not so much.
I think 0.44N was not such a great statement, but I know what you meant [it reads like a force value in Newtons]

Next I am not sure why you decided to divide by 0.44 - nor why you thought the answer was what you gave.

I don't think you should round of 4/9 to 0.44, and think more carefully about how to use that factor.
 
  • #3
PeterO said:
Up to (3/4.5)^2 I was fairly comfortable with.

From there on, not so much.
I think 0.44N was not such a great statement, but I know what you meant [it reads like a force value in Newtons]

Next I am not sure why you decided to divide by 0.44 - nor why you thought the answer was what you gave.

I don't think you should round of 4/9 to 0.44, and think more carefully about how to use that factor.

Yes good idea, i'll stick with 4/9 instead of the decimal to preserve accuracy. Since F is proportional to 1/R^2. If R increases from 3 m to 4.5 m shouldn't the force decrease by a factor of (3/4.5)^2. And i thought decrease by a factor means divided by so i divide the original force by the force i got from the divide i did prior?

Edit: I just divided it the 4/9, it gives me 6.9x10^-8 ( i see why you suggested i'd use 4/9 )
 
  • #4
xNick94 said:
Yes good idea, i'll stick with 4/9 instead of the decimal to preserve accuracy. Since F is proportional to 1/R^2. If R increases from 3 m to 4.5 m shouldn't the force decrease by a factor of (3/4.5)^2. And i thought decrease by a factor means divided by so i divide the original force by the force i got from the divide i did prior?

Edit: I just divided it the 4/9, it gives me 6.9x10^-8 ( i see why you suggested i'd use 4/9 )

It is common for people to confuse what to do with the factors.

You calculated a factor of 4/9 [good], and wanted to find the reduced force. That is, you wanted to get a smaller value for the Force.

How do you use 4/9 to get a smaller value?

Note: Some people get a factor of 9/4 [perhaps in error]. what should they do with a factor of 9/4 in order to get a smaller value?

And don't forget to apply your factor to the original value given!
 
  • #5
PeterO said:
It is common for people to confuse what to do with the factors.

You calculated a factor of 4/9 [good], and wanted to find the reduced force. That is, you wanted to get a smaller value for the Force.

How do you use 4/9 to get a smaller value?

Note: Some people get a factor of 9/4 [perhaps in error]. what should they do with a factor of 9/4 in order to get a smaller value?

And don't forget to apply your factor to the original value given!

Hmm so was my original divide procedure justified from my explanation?
3.1x10^-8 / (4/9) = 6.9x10^-8 ...the force increased so this has to be wrong since it has to decrease right?
 
  • #6
xNick94 said:
Hmm so was my original divide procedure justified from my explanation?
3.1x10^-8 / (4/9) = 6.9x10^-8 ...the force increased so this has to be wrong since it has to decrease right?

Yes - it is supposed to decrease.

I never bother too much which factor my students calculate [4/9 or 9/4], I just want them to think about how to effect the change needed.
If you want a smaller answer, and will be using a factor less than 1, you need to multiply.
If you want a smaller answer, and will be using a factor greater than 1, you need to divide.

Send me back your new answer then I will explain how I would have looked at/solved the problem.
 
  • #7
xNick94 said:
Hmm so was my original divide procedure justified from my explanation?
3.1x10^-8 / (4/9) = 6.9x10^-8 ...the force increased so this has to be wrong since it has to decrease right?

Did you sort out where you went wrong? Only a single, simple error!
 
  • #8
Sorry for the late reply, had to work and family friends were visiting.

If you want a smaller answer, and will be using a factor less than 1, you need to multiply.
If you want a smaller answer, and will be using a factor greater than 1, you need to divide.

PeterO said:
Did you sort out where you went wrong? Only a single, simple error!

Would you then have to multiply it since 4/9 is under 1.
3.1x10^-8 x 4/9 = 1.3 x 10^-8

Would the force be just 1.3x10^-8 or would i have to subtract that from 3.1x10^-8?
 
  • #9
xNick94 said:
Sorry for the late reply, had to work and family friends were visiting.

Would you then have to multiply it since 4/9 is under 1.
3.1x10^-8 x 4/9 = 1.3 x 10^-8

Would the force be just 1.3x10^-8 or would i have to subtract that from 3.1x10^-8?

The multiplication is correct. Though you have merely "truncated" the answer rather than round it off - it should be 1.4 x 10^-8 when rounded to 2 figures.

Fortunately if you had subtracted 4/9 form 3.1 x 10^8 you would get approximately -4/9 and thus recognise it as wrong.

4/9 was a factor, and you either multiply or divide by factors.

Fractions are always a problem, because often work backwards: multiplying by them you get smaller, divide you get bigger.

The way I would have done it - which ultimately is numerically the same thing - is.

Since separation is changing from 3 to 4.5 - not an integer factor change - I would work on an interim separation of 1.5 m

This is half the first separation - meaning 4 x the force or 12.4 x 10^-8

I then change to 4.5.

This is 3 times the interim separation so it if F / 9 or 1.377778 x 10^-8

Rounding to 2 significant figure this is 1.4 x 10^-8

********** Be careful to always do calculations to as many figures as your calculator can handle - then at the end ROUND OFF, don't just chop off the last figures.**************
 

1. What does the equation F = kQq/R2 represent?

The equation F = kQq/R2 represents Coulomb's Law, which describes the electrostatic force between two charged objects.

2. How do you find the value of k in the equation?

The value of k, also known as the Coulomb's constant, can be found by dividing the permittivity of free space (ε0) by 4π. Its value is approximately 8.99 x 10^9 Nm^2/C^2.

3. What do Q and q represent in the equation?

Q and q represent the magnitudes of the two charges in the system. Q is the charge of the first object and q is the charge of the second object.

4. How do you calculate the force between two charged objects using this equation?

To calculate the force between two charged objects, you need to know the values of k, Q, q, and the distance between the two objects (R). Then, simply plug these values into the equation F = kQq/R2 and solve for F.

5. What is the unit of measurement for the force calculated using this equation?

The unit of measurement for the force calculated using this equation is Newtons (N), which is a unit of force in the International System of Units (SI).

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
875
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top