The force when the distance is reduced

[aChordate]

Homework Statement

Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

F=1.5 N

Homework Equations

F=k|q1||q2|/r^2

The Attempt at a Solution

1.5 N = k |q1||q2|/(1/9)r^2

I don't think I am setting this up right, but would I multiply 1.5N by 1/9 and by the constant k?

Thanks in advance!

 

[Doc Al]

I don't think I am setting this up right, but would I multiply 1.5N by 1/9 and by the constant k?

Set it up so that 1.5 N is the force when the distance = R.

Then see what the force is when the distance = R/9.

(Set up a ratio.)

 

[Doc Al]

Another way to approach this is to just ask yourself what happens to the force if the distance doubles? If the distance is cut in half? That should point you in the right direction.

 

[aChordate]

So,

1.5N/1=F/(1/9)

then,

F=0.17 ?There is an identical homework problem with the solution on this page (http://kingfish.coastal.edu/physics/phys202_S05/HW/HW_18.pdf) , but I don't understand what they did to get the answer.

 

[Doc Al]

Try answering the questions I asked in post #3.

 

[aChordate]

Another way to approach this is to just ask yourself what happens to the force if the distance doubles? If the distance is cut in half? That should point you in the right direction.

If the distance doubles, the force increases x4 (exponentially?). If the distance is halved, the force decreases.

 

[Doc Al]

If the distance doubles, the force increases x4 (exponentially?).

Careful. The force is inversely proportional to the distance squared. So if the distance doubles, the force decreases by a factor of 4. (Multiplied by (1/2)² = 1/4.)

If the distance is halved, the force decreases.

No, the force would increase by a factor of 4.

F₁ = (const)*(1/R)²

If you reduce the distance by 2, replace R with R/2:

F₂ = (const)*(1/(R/2))² = 4*(const)*(1/R)² = 4*F₁

(Now do the same thing for the distance being reduced by 1/9.)

 

[aChordate]

F2 = (const)*(1/(R/9))2 = 81*(const)*(1/R)2 = 81*1.5 N = 121.5 N

Wow, I really need to work on more of these.

Thanks so much for your help!

 

[cathal84]

Hi Doc Al,

I am trying to understand the answer that you posted above.

F2 = (const)*(1/(R/2))2 = 4*(const)*(1/R)2 = 4*F1

Could you explain why the squared sign is outside the last bracket. Should it not be inside the bracket as it is the distance squared.

F2 = (const)*(1/(R/2)2)

Also, how did you simplify it to be 4*(const), how did you calculate the 4?

Many thanks,
Cathal

 

[Doc Al]

Could you explain why the squared sign is outside the last bracket. Should it not be inside the bracket as it is the distance squared.

Realize that $(\frac{1}{R})^2 = \frac{1}{R^2}$

Also, how did you simplify it to be 4*(const), how did you calculate the 4?

$(\frac{1}{R/2})^2 = (\frac{2}{R})^2 = 4(\frac{1}{R})^2$