## anonymous 3 years ago Two point charges of equal magnitude are 7.5 cm apart. At the midpoint of the line connecting them, their combined electric field has magnitude of 45 N/C. Find the magnitude of the charges. Pls show me the complete calculation

1. anonymous

first tell me what is the formula to calculate electric field due to a point charge?

2. anonymous

E=F/q=kq/r^2

3. anonymous

ok.. good.. tell me.. do you think the charges should be of same nature or opposite nature?? or both possible? doesn't matter?

4. anonymous

i think it has the same nature since the magnitude of electric field is high

5. anonymous

ok.. if the charges are of same nature.. then think the direction of the electric field due to each charge at the mid point? will they be same direction or opposite direction?

6. anonymous

|dw:1354890053415:dw|

7. anonymous

i think same direction

8. anonymous

i dont know what happens at the center :-/ the radius is halved i guess?

9. anonymous

how is that?? the direction of electric field due to a positive charge is always AWAY from the charge!!

10. anonymous

|dw:1354890336186:dw|

11. anonymous

okay

12. anonymous

|dw:1354890474542:dw|

13. anonymous

now.. Ea + Eb = the required answer.. !!... Ea you can find.. Eb you can find.. add them equate to required answer.. get Q

14. anonymous

the net force is 0?

15. anonymous

i did try that but i cant seem to get the correct anwers. the answer is 3.5 x 10^-12

16. anonymous

No now.. the net force will add up and = 45 N/C

17. anonymous

net field*

18. anonymous

$2Q/4\pi \epsilon r ^{2}= 45$

19. anonymous

r is .075/ 2 cause the mid point is half the distnace

20. anonymous

i got the answer! thank you very much :)

21. anonymous

your welcome!.. i could have given the answer before.. but i wanted you to think and get it :P

22. anonymous

i know it helps a lot! so thank you!

23. anonymous

anytime :)!