# Homework Help: Point charges

1. Jul 14, 2010

### pat666

1. The problem statement, all variables and given/known data

2. Relevant equations

F=kq1q2/r^2

3. The attempt at a solution
I cant see where to start this question, I can do questions where there is three point charges and it asks you to find the net force on one of them and i think that in essence this question is the same. the problem is that nothing i do/think of seems to work. I thought that F1x + F2x =0 might work but i always end up with two variables (r and qx). Please help

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2. Jul 14, 2010

### Ush

1. Forces are vectors. Forces in the same direction can be added together.
All three charges lie on the x-axis, including your unknown charge (why?, hint: draw diagrams showing electric field vectors of positive charges)

2. what is the only region where you can put this charge so the forces can potentially add up to zero?

3. Jul 14, 2010

### pat666

at some point in the middle im thinking but i was trying to find the co-ordinates - is that not what part a is after???? if so I still cant see how to do part b. thanks ush

4. Jul 14, 2010

### Ush

yes.
You know the total distance between the two charges is 0.9m
let "a" be the distance x=0 to the unknown charge
now, without adding another variable, what is the distance from the unknown charge to x=0.9 ?

5. Jul 14, 2010

### pat666

how can i work that out???????- sorry but i am now thoroughly confused....

6. Jul 14, 2010

### Ush

lol.
if 'a' is the distance from x=0 to the unknown charge, then '0.9 - a' must be the distance from the unknown charge to x=0.9

why?

if you add "a" + "0.9 - a" you get your total distance "0.9". so 'a' is somewhere within your total distance.

knowing this.. you can plug in these variables for your radius. ("a" being the radius from 5.0 μC to the unknown charge, and "0.9-a" being the radius from the 8.0 μC to the unknown charge).

solving for "a" is now strictly algebraic
--
Recall: the two forces exerted on the point cancel out, therefore the force exerted on the unknown point by the 5.0 μC charge cancels with the 8.0 μC charge on your free body diagram (which I highly recommend drawing at the very beginning).

7. Jul 14, 2010

### pat666

thankyou!!!!! Ill post back if i have any more issues- thanks!!!!!!!!!!!!!!!

8. Jul 14, 2010

### pat666

hey ush, ive been trying to solve this for ages now and i keep getting -1.5m and that cant be right. I am not sure what i am doing wrong??

9. Jul 14, 2010

### pat666

if i minus them instead of +ing them i get 0.397m??? - i really need help..

10. Jul 15, 2010

### graphene

Put down your solution here. I'll check it.

11. Jul 15, 2010

### pat666

thanks graphene
F1x+F2x=0
0=9E9*5E-6/a^2+-9E9*8E-6/(0.9-a)^2
then I solve for a using the 89 and get 0.397 or -1.5 depending on plus or minus in the middle

12. Jul 15, 2010

### graphene

So, a = 0.397 is the only possible solution.

That gives you the the position.

Now you'll have to find out what charge you need to place there so that the net force on the first two charges is zero.
What should be the sign of the third charge so that the net force is zero on the first two charges after the third charge is placed?

13. Jul 15, 2010

### Ush

-1.5 is a distance that's NOT in-between the two charges, 0.397 is between the two charges. You concluded earlier that the unknown charge must be in-between the two charges.

14. Jul 17, 2010

### pat666

Hey ush sorry for the late reply- I have solved the problem - just wondering if you could check it please... I got d=0.397m and Q_3=-1.54*10^(-6) C

15. Jul 17, 2010

### Ush

Your distance is correct.
You're charge for Q3 is incorrect. Q3 can be any charge it wants to be.. it does not matter because on the free body diagram of Q3 the only forces acting on it are Q1 and Q2. Q3 cannot exert a force on itself therefore it makes no difference what Q3 is.

Unless I'm missing something from the question- I don't believe Q3 has to be any particular magnitude of charge.

16. Jul 17, 2010

### pat666

hey Ush - if you look at question b it does want both the magnitude and sign

17. Jul 18, 2010

### graphene

Q3 cannot be any charge.
The questions requires that the forces on q1 & q2 be zero as well. So, you'll need to place a particular charge for q3.

pat666 - again, if you could show me your solution, i'll verify it.

18. Jul 18, 2010

### pat666

I got d=0.397m and Q_3=-1.54*10^(-6) C
thanks again graphene very very helpful

19. Jul 18, 2010

### Ush

You're right, I misread the question. I thought it was referring to Q3 only, but it says all charges.

In that case, a charge of -1.55E-6 should balance the forces, so yeah, that's right

20. Jul 18, 2010

### pat666

Thanks for all your help