## woleraymond 3 years ago Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm on the x axis. A third charge e is placed on the straight line joining the two charges so that e is in equilibrium. Find the position of e.

1. woleraymond

@Algebraic! any clues

2. Algebraic!

|dw:1348695082199:dw|

3. Algebraic!

q1 (+4e) pushes a positive charge to the right. q2 (+2e) push a positive charge to the left. at some point between them those opposing pushes will be equal in magnitude. sub.s in x and .4-x for r1 and r2 and solve for x

4. Algebraic!

is it clear?

5. woleraymond

not clear at all

6. Algebraic!

the forces are equal in magnitude and opposite in direction.

7. Algebraic!

| e*E1 |=| e*E2 | E1 = kq1/(r1)^2 E2 = kq2/(r2^2)

8. Algebraic!

r1 is the distance from the charge at the origin to the location of the point where they 'test' charge is in equilibrium. That distance is x. r2 is the distance from the charge at the .4m to the location of the point where they 'test' charge is in equilibrium. That distance is .4m - x

9. Algebraic!

E1 = kq1/(r1)^2 = E2 = kq2/(r2)^2 kq1/(r1)^2 = kq2/(r2)^2 kq1/(x)^2 = kq2/(.4-x)^2

10. Algebraic!

11. Algebraic!

clear now?

12. woleraymond

@Algebraic! not clear..how do you interepret Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm..also how u derive the value for r1 and r2..what about the 3rd charge?

13. Algebraic!

+4e charge is at x=0; +2e charge is at x=.4

14. Algebraic!

r for the +4e charge is the distance from x=0 to the equilibrium point we're trying to find... call that distance x, r for the +2e charge is then the distance from x=.4 to the equilibrium point we're trying to find so .4-x

15. Algebraic!

|dw:1348774582953:dw|

16. woleraymond

there is no 4 its 40..also the (0,0) isnt the x,y pairs?

17. Algebraic!

.4

18. Algebraic!

both charges are on the x axis and the equilibrium point is on a line between them, so it's on the x axis too. I don't know why they wrote it like this: "(0,0)cm and (0,40)cm on the x axis." it contradicts itself. so maybe check it to see if that's what the problem actually says. I assumed what they actually meant was both charges on the x axis one at x=0 one at x=.4

19. Algebraic!

anyway, it's the same problem if those are actually meant to be ordered pairs, because: |dw:1348775056115:dw|

20. Algebraic!

ok?

21. woleraymond

so u mean the 3rd charge is in the middle of both charge?

22. Algebraic!

you have to locate where the third charge is... ...that's the point of the problem. but yes, it's probably on the x-axis, depending on what your problem statement actually says...

23. woleraymond

how do we solve this kind of quadratic 4e/x^2 = 2e/(4-x)^2

24. Algebraic!

4e/x^2 = 2e/(.4-x)^2

25. Algebraic!

cross multiply

26. Algebraic!

4e(.4-x)^2 = 2ex^2

27. woleraymond

x is 2.3

28. woleraymond

@Algebraic! do you believe i just saw the .4 and not 4...thanks a lot