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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Use equation, E=F/q
what i have is \[E=1/(4\pi \epsilon _{0}) * Q/r^{2}\]
but you don't have charge. so you can't use this relation.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[Q=L \lambda\] and the prof said use 2 for L. not sure why, maybe 10cm/5cm but why do that?
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
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
did you got?
why L=2?? may be due to doubling the length.
not really, i can't recall how to solve for force. force of field on wire or vice versa?
or are you saying solve for force after i find E?
ok follow me.
alright
use 1 equation and find Q by substituting values, k=9*10^9 E=2000 r=0.05m
right
Q=5.56*10^-10
ok good. Now use equation 2 and find F. by substituting values of, E=2000 q=5.56*10^-10
should be 1.11*10^-6
i will try to work more on it you will get a notification if i add more to this later, gotta go, thank you
you are welcome.
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 \( r = 0.10 m \).
hey jamesj who is this person who is reviewing 3rd question of emcrazy14?
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...
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
thx, i had my r squared, should be able to get it now

Not the answer you are looking for?

Search for more explanations.

Ask your own question