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## anonymous 4 years ago two 10cm diameter charged rings face each other 20cm apart. both rings are charged to +40nC. what is the electric field strength...

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1. anonymous

a) at the midpoint between 2 rings? should the E just be added together but using r=10cm?

2. EarthCitizen

$E=F/Q$

3. anonymous

found the answer on cramster step by step but not understanding the r: Electric field due to charged ring E = KQx / [ x ^ 2 + R^2 ] ^ 1.5

4. anonymous

(x^2+R^2)^1.3??

5. anonymous

oops 1.5

6. EarthCitizen

is it between to plates ?

7. anonymous

yes

8. EarthCitizen

two plates*

9. anonymous

yes

10. EarthCitizen

\[\epsilon _{0}\ = ?]

11. anonymous

8.85*10^-12

12. anonymous

k=1/(4pi8.85*10^-12)=8.99*10^9

13. EarthCitizen

first convert cm to m

14. anonymous

right

15. anonymous

.1m diamter, .2m apart

16. anonymous

and +40nC=+

17. anonymous

+40*10^-9

18. anonymous

answer says it should be 0 cuz they cancel each other but i thought it would double since both rings are +

19. EarthCitizen

both rings are either + /-

20. EarthCitizen

calculate the force, r= d/2

21. anonymous

i actually understand now from the set up on cramster except for the part that is raised to 1.5...typo? should be .5

22. anonymous

o charge density for ring=Q/(x^2+r^2)^1.5

23. anonymous

i mean Q/(2pi*r)

24. EarthCitizen

r=d/2

25. EarthCitizen

d=0.1m/2 = 0.05m

26. EarthCitizen

total net E=E1+E2

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