Question

A ring of radius a contains a charge q distributed uniformly over its length.Find the electric field a point on the axis of the ring at a distance x from the centre

Answer 1

@shubhamsrg @yrelhan4

Answer 2

what are you going to use?

Answer 3

coloumbs law or

Answer 4

Gauss?

Answer 5

coulumbs

Answer 6

hmm let me think about you'd do this

Answer 7

\[dq=\lambda dx\]
where
\[\lambda\]is the linear charge density

Answer 8

now what i'm thinkins that since it's a symmetrical shape such as a circle the vector on each side will cancel everything out leaving only a horizontal component of E named

\[\E_x\]

Answer 9

i know that it is for a uniform line of charge and as it looks right now... the only vectors that do not cancel is the horizontal ones

Answer 10

so if

\[E=\frac{kQ}{r^2}\]

Answer 11

\[dE=\frac{kdQ}{r^2}=\frac{k\lambda dx}{r^2}\]

Answer 12

now hmm gotta think of something for r^2

Answer 13

\[\LARGE E_x=\frac{KQ}{(R^2+a^2)^{3/2}}\]
I got this.

Answer 14

that looks right

Answer 15

Nope thats not right. You have an 'a' missing in the numerator.

Answer 16

oh hmm

Answer 17

oh sorry :|

Answer 18

\[\LARGE dE_x=\frac{dQ}{4\pi \epsilon_o \cos \theta}\]