# Homework Help: Can someone help a newbie, electric forces

1. Jul 20, 2004

### DSM4Life

Ok this is the question in text book i cant 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

2. Jul 20, 2004

### Staff: Mentor

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.

3. Jul 20, 2004

### DSM4Life

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

4. Jul 20, 2004

### Staff: Mentor

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.

5. Jul 22, 2004

### jatin9_99

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

6. Jul 22, 2004

### 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 $$\frac{ F_{new} }{ F_{old} } = \frac{1}{5^2}$$.