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

Question

shayaan_mustafa · Answer

Use equation,
E=F/q

anonymous · Answer

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

shayaan_mustafa · Answer

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

anonymous · Answer

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

anonymous · Answer

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

shayaan_mustafa · Answer

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

shayaan_mustafa · Answer

did you got?

shayaan_mustafa · Answer

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

anonymous · Answer

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

anonymous · Answer

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

shayaan_mustafa · Answer

ok follow me.

anonymous · Answer

alright

shayaan_mustafa · Answer

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

anonymous · Answer

right

anonymous · Answer

Q=5.56*10^-10

shayaan_mustafa · Answer

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

anonymous · Answer

should be 1.11*10^-6

anonymous · Answer

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

shayaan_mustafa · Answer

you are welcome.

jamesj · Answer

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 $.

shayaan_mustafa · Answer

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

anonymous · Answer

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...

jamesj · Answer

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

anonymous · Answer

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