# Homework Help: Physics Problem Help

1. Oct 27, 2004

### suf7

Physics Problem!!..Help!!

Can sum1 pleez help me with this problem???...i really dont have a clue???...i dont unsterstand what is the correct method to use???...need help!!

Three point charges are situated on the x-axis of a co-ordinate system.
Q1 = +2.0 nC is +0.05m from the origin
Q2 = -3.0 nC is + 0.1m from the origin
Q3 = -10.0 nC is -0.1m from the origin
What is the total force exerted by the three charges on a fourth charge “Q4 = +5.0 nC is situated at the origin??.....1 nC = 10-9C

2. Oct 27, 2004

### arildno

So you must find out:
1) What is the magnitude and direction of the force from Q1 on Q4?
2) What is the magnitude and direction of the force from Q2 on Q4?
3) What is the magnitude and direction of the force from Q3 on Q4?
4) Then add the results from 1),2),3) together
5) Remember:
Whatever forces Q1,Q2,Q3 imparts ON EACH OTHER has NO RELEVANCE to the forces they impart on Q4!!!

3. Oct 27, 2004

### Nylex

Use Coulomb's law to find the force acting on Q4 exerted by each of the other charges. The total force exerted on Q4 is given by the sum of those 3 forces.

4. Oct 27, 2004

### suf7

Sorry but im really lost???...could you show me how to work out the first part???
What is the magnitude and direction of the force from Q1 on Q4?

ive never been shown any method for working these questions out, my teacher jus gives us the answers???...i just need to know what steps to take and how to achieve an answer????

5. Oct 27, 2004

### arildno

6. Oct 27, 2004

### suf7

force between two stationary point charges???

7. Oct 27, 2004

### arildno

Precisely!

8. Oct 27, 2004

### suf7

i know thats the law, i just dont know how to apply it???

9. Oct 27, 2004

### arildno

Okay then, let's start at my 1)

Set up Coulomb's law to calculate the force from Q1 on Q4!

10. Oct 27, 2004

### suf7

but i dont know how to???..i been given some formulas but i dont know which is which and which1 to use??..im confused??..sorry!!

11. Oct 27, 2004

### arildno

Can you type in Coulomb's law, using Q1 and Q4 as your point charges?

12. Oct 27, 2004

### drnikitin

Find forces (magnitude and direction) from each of the three charges on the fourth charge from Coulomb's law. Then add these forces taking into account their directions

13. Oct 27, 2004

### suf7

is it this??

F = (Q1)(Q4)/4∏eor2

14. Oct 27, 2004

### suf7

by the way how does nC come into all this???

15. Oct 27, 2004

### Nylex

nC = nano Coulombs. Charge is measured in Coulombs, so you need to convert your values into Coulombs (ie. multiply by 10^-9). The equation you have above for Coulomb's law was correct (assuming you mean r^2 on the bottom line).

16. Oct 27, 2004

### arildno

That is correct, as long as you remember that the implied direction is AWAY from Q4.
On vector form, we have:
$$\vec{F}_{41}=\frac{1}{4\pi\epsilon_{0}}\frac{Q_{1}Q_{4}}{r_{41}^{2}}\frac{\vec{r}_{4}-\vec{r}_{1}}{r_{41}}$$
EXPLANATIONS:
1)Here, I have included the last fraction to have the basic direction explicitly included.
2)$$\vec{F}_{41}$$ means: The force acting on Q4 from Q1
3) $$r_{41}$$ is the DISTANCE between Q1 and Q4, that is, some positive number.
CALCULATIONS:
Q1=2.0 nC
Q4=5.0 nC
Hence, $$Q1*Q4=10.0(nC)^{2}$$
$$\vec{r}_{4}=\vec{0}$$ (that is, situated at the origin)
$$\vec{r}_{1}= 0.05m\vec{i}$$
(that is, situated 0.05m to the right-hand side of Q4 at the origin)
Hence, $$\vec{r}_{4}-\vec{r}_{1}=-0.05m\vec{i}$$
That is, the direction indicated by this quantity is leftwards.

$$r_{41}=0.05m$$ that is the DISTANCE between Q1 and Q4 is 0.05m

Collecting all together, we have:
$$\vec{F}_{41}=-\frac{1}{4\pi\epsilon_{0}}\frac{10.0(nC)^{2}}{(0.05m)^{2}}\vec{i}$$

That is, the force on Q4 from Q1 is directed leftwards.
Q4 tends to be REPELLED from Q1, because they have EQUAL TYPE OF CHARGE!

17. Oct 27, 2004

### suf7

So is this right for the first point??

F = (Q1)(Q4)/4∏eor2
= (2*5) / (4∏)*(8.854*10^-12)*(2.5*10^-21)
= 3.597*10^31

18. Oct 27, 2004

### suf7

thanks for that explanation..sorry, im abit slow..lol..
But wats hapend to the 4pi??and Eo??..wat have i done wrong in my calculation???

19. Oct 27, 2004

### Nylex

Why are you using 2.5 x 10^-21 as the distance between the charges? Also remember to convert your charges into coulombs if you want your force in newtons.

20. Oct 27, 2004

### arildno

You should have (I include the sign (direction) for the force):
$$F=-\frac{10.0*10^{-18}C^{2}}{4\pi*\epsilon_{0}*2.5m^{2}*10^{-3}}$$
If I understand your calculation correctly, you placed the $$10^{-18}$$ factor in your denominator, rather than in your numerator.