If I have a charged ring of radius $a$ what I'm trying to find the electric field of a point $r$ from the axis in the plane for $r\ll a$. The example sheet hints to use a gaussian surfaces of radius $r$ but I'm not sure what purpose a gaussian surface that encloses no charge serves and so i've tried to to is directly.
By symmetry the only component of the field will be radial. Using Coulomb's law in conjunction with the cosine rule I get
$$E=\frac {\sigma} {4\pi \epsilon_0} \int_{0}^{2\pi} {\frac {r-a\cos(\theta)}{((a^2+r^2)-2ar \cos(\theta))^{3/2}} \ d\theta}$$
Is this integral possible using first year university maths?
Is there an easier way to find the electric field off axis for $r\ll a$ in the plane of a charged ring?
Best Answer
So if we take a very small gaussian pillbox centred on the origin of height $2z$ and radius $r$ in the limit the field out of the top and bottom surfaces is: $$\frac {Qz\pi r^2}{4\pi\epsilon_0(a^2+z^2)^{3/2}}$$ therefore as the total charge enclosed is zero and we know the field through the sides of our pillbox is radial and of constant magnitude we can arrive at: $$ \frac {2Qz\pi r^2}{4\pi\epsilon_0(a^2+z^2)^{3/2}} - 4\pi rzE = 0$$ This rearranges to: $$E=\frac{Qr}{8\pi\epsilon_0(a^2+z^2)^{3/2}}$$ which is: $$\frac{Qr}{8\pi\epsilon_0a^3}$$ if we take the taylor series and say the $\frac{z^2}{a^2}$ and above terms are negligible in the limit $z<<a$. Seems to be quite wishy washy but it does get the right answer.
Thanks to wltrup for help!