anonymous 4 years ago the electric field strength 5cm from a very long charged wire is 2000N/C. what is the electric field strength 10cm from the wire?

1. Shayaan_Mustafa

Use equation, E=F/q

2. anonymous

what i have is $E=1/(4\pi \epsilon _{0}) * Q/r^{2}$

3. Shayaan_Mustafa

but you don't have charge. so you can't use this relation.

4. anonymous

$Q=L \lambda$ and the prof said use 2 for L. not sure why, maybe 10cm/5cm but why do that?

5. anonymous

after i solve for 5cm, i use the value i got for charge density to plug in for 10cm and got 500 N/C. i don't understand why L=2 and if i still use L=2 to solve for 10cm

6. Shayaan_Mustafa

there are two equations, 1) E=kq/r² find charge from above equation and put it in, 2) E=F/q find Force from above relation

7. Shayaan_Mustafa

did you got?

8. Shayaan_Mustafa

why L=2?? may be due to doubling the length.

9. anonymous

not really, i can't recall how to solve for force. force of field on wire or vice versa?

10. anonymous

or are you saying solve for force after i find E?

11. Shayaan_Mustafa

12. anonymous

alright

13. Shayaan_Mustafa

use 1 equation and find Q by substituting values, k=9*10^9 E=2000 r=0.05m

14. anonymous

right

15. anonymous

Q=5.56*10^-10

16. Shayaan_Mustafa

ok good. Now use equation 2 and find F. by substituting values of, E=2000 q=5.56*10^-10

17. anonymous

should be 1.11*10^-6

18. anonymous

i will try to work more on it you will get a notification if i add more to this later, gotta go, thank you

19. Shayaan_Mustafa

you are welcome.

20. JamesJ

The electric field about a long, straight charged wire is given by $E(r) = \frac{\lambda}{2\pi \epsilon_0 r}$ where $$\lambda$$ is the charge density. (See for example here: http://farside.ph.utexas.edu/teaching/302l/lectures/node26.html ) Given that, and what you are told about $$E(r)$$ for $$r = 0.05 m$$, you can then find $$E(r)$$ for $$r = 0.10 m$$.

21. Shayaan_Mustafa

hey jamesj who is this person who is reviewing 3rd question of emcrazy14?

22. anonymous

i am still confused. i solved for the charge density given 2000N/C for 5cm but i used 2 for L. L isn't given so i really don't understand why 2. well, it's the same wire the question is asking about so i used it and my answer is 500N/C which seems reasonable...

23. JamesJ

From the formula I wrote above for the electric field, you see that it scales as 1/r. So at twice the distance away it should be 1/2, not 1/4

24. anonymous

thx, i had my r squared, should be able to get it now