Question

A circular wire loop of radius 'a' carries a total charge 'Q' distributed uniformly over its length. A small length dL of the wire is cut off. What is the electric field at the center due to the remaining wire?

Answer 1

the field that will act will be that for which there is no side symmetrically opposite it. That is, the other side that was cut out

Answer 2

\[dQ=\lambda dL\]where \(\lambda\) is the linear charge density of the wire. Coulombs law gives\[d\vec E=k{\lambda\over a^2}dL\hat a\]

Answer 3

U r right @TuringTest :)
But the answer given in my book is different & that is \[\huge{\color{red}{\vec E = \frac{Q \space dL}{8 \pi^2 \epsilon_0a^3}.}}\]

Answer 4

How can i get this answer ?
Please anyone help:)

Answer 5

@experimentX Please help:)

Answer 6

doesn't look easy

Answer 7

ya:(

Answer 8

Am I Right ?\[\LARGE{\vec E=\frac{\frac{Q}{2 \pi a}\times dL}{4 \pi \epsilon_0 \space a^2}}= \frac{Q \space dL}{8 \pi^2 \epsilon_0a^3}.\]

Answer 9

Electric field due to a line element is
\[ E = { k \lambda dl \over r^2} \\
l = 2 \pi r \implies dl = 2 \pi dr\]

Answer 10

seems so ...

Answer 11

I have only thought about that.
I think I must confirm this.

Answer 12

probably it's the easiest way to do it.

Answer 13

\[ E = { k \lambda dl \over r^2} \\
\lambda = {q\over 2 \pi r}\]

Answer 14

yea so i m right:)

Answer 15

i had been thinking the straight way when the back way was more straight forward

Answer 16

hmm......

Answer 17

May i close this question?
i mean
have we gt the solution?

Answer 18

yeah ...

Answer 19

ok.
thank u very much:)

Answer 20

originally ... the E at center is zero. it means that the E due to rest must have been balanced by dl ... so E due to dl should be equal to E due to rest.

Answer 21

ya .

Answer 22

@experimentX i have a confusion.
would u plz help me ?

Answer 23

where do you have?

Answer 24

I was saying that
if i wanna have charge of remaining wire then i can subtract dL part's charge from net charge Q.
Then
can i get electric field due to the remaining wire?

Answer 25

@experimentX ?

Answer 26

where did my answer go?
dL will be equal to remaining part.

Answer 27

but i m asking that
can't i get this way?

Answer 28

yes you can ..

Answer 29

but actually i m nt getting
would u plz show me?

Answer 30

\[ Q - {Q \over 2 \pi a} dL\]

Answer 31

ya I did this but this is wrong:(\[\LARGE{\vec E=\frac{Q-\frac{Q}{2 \pi a}\times dL}{4 \pi \epsilon_0 \space a^2}}= \frac{2 \pi a Q -Q \space dL}{8 \pi^2 \epsilon_0a^3}.\]

Answer 32

It must be \[\huge{\color{red}{\vec E = \frac{Q \space dL}{8 \pi^2 \epsilon_0a^3}.}}\]
Please help:)

Answer 33

I have problem in numerator.

Answer 34

what are you trying to do? are you trying to get it the other way?

Answer 35

ya.

Answer 36

I have gt the remaining wire's charge so that i can get electric field due to that wire.
but i m getting wrong:(