Can someone help a newbie, electric forces

[DSM4Life]

Ok this is the question in textbook i can't figure out.
"The force of repulsion that two like charges exert on each other is 3.5N. What will the force be of the distance between the charges is increased to fice times its orginal value ? "

im a physics moron , treat me like one :rofl:

ok i understand that two like charges will repluse.
In this case its at 3.5N. So i was thinking that if the force of it was at 3.5N and the distance was increased by 5x then just multiply 3.5N x 5 = 17.5 which when i check in back off the book is wrong. Can you help explain this to me so i can understand it ? The answer the book has is 0.14N

 

[Doc Al]

Coulomb's law

Welcome to PF!

You'll need to understand Coulomb's law, which describes the force between two charges:
$$F = k\frac{q_1q_2}{r^2}$$
Note that the force decreases as the charges get further apart, but it's proportional to 1 over the square of the distance. For example: if the distance is doubled, the force is 1/2 x 1/2 = 1/4 as much--four times smaller.

 

[DSM4Life]

Welcome to PF!

You'll need to understand Coulomb's law, which describes the force between two charges:
$$F = k\frac{q_1q_2}{r^2}$$
Note that the force decreases as the charges get further apart, but it's proportional to 1 over the square of the distance. For example: if the distance is doubled, the force is 1/2 x 1/2 = 1/4 as much--four times smaller.

see i was looking at that in my book. The thing that was throwing me off is that i thought i needed another q value.

I understand its the constant K then q1 point and q2 divided by sq distance but i didnt know where the second value was. Still little confused.

F = (8.99x10^9) |q1||q2|
---------
r^2

umm what is the distance and points

 

[Doc Al]

You are given everything you need to answer the question. You don't need to know the charges (they don't change) and you don't need to know the actual distance--just by what factor it increases. If you understand what Coulomb's law is telling you--you can solve this problem just by ratios.

As I said before: if the distance doubles, then the force decreases by a factor of four. Right? And if you know what the force was to start with, then you can find out what it ends up being. Right?

Now in your problem the distance increases by a factor of 5. So by what factor does the force decrease? Figure out that new force.

 

[jatin9_99]

the new force would be i/25 th of the original force according to coloumbs law f=kq1q2/r^2

 

[robphy]

Here's a way to think about what Doc Al said:

Given $$F = k\frac{q_1q_2}{r^2}$$ and $$F=3.5\ N$$,
you want $$F_{new}$$ when $$r_{new}=5r_{old}$$.



Observe
$$\begin{align*} F_{new} &= k_{new}\frac{q_1_{new}q_2_{new}}{r_{new}^2} && \text{Coulomb}\\ &= (k_{old})\frac{(q_1_{old})(q_2_{old})}{(\textcolor{red}{5}r_{old})^2} && \text{new in terms of old}\\ &= \frac{1}{\textcolor{red}{5}^2}(k_{old})\frac{(q_1_{old})(q_2_{old})}{(r_{old})^2} && \text{algebra}\\ &= \frac{1}{\textcolor{red}{5}^2} F_{old}&& \text{Coulomb}\\ \end{align *}$$

Note the ratio $$\frac{ F_{new} }{ F_{old} } = \frac{1}{5^2}$$.