A circular copper ring of radius r, placed in vacuum, has a charge q on it. Find out the electric fields on the axis of the ring at a distance x from its centre. ?

Explanation

The charge q would be uniformly distributed along the circumference of the circular ring. So, the linear density of charge along the ring.
`lamda=q/(2pir)`
Let us take a small element of length dl on the ring at maximum the position A [Fig.2.28]. Charge on the element `dl = lamdadl=q/(2pir)dl`. The electric field at the axial point P due to the charge on the element dl,

`dE=1/(4iepsilon_(0)).(lamdadl)/((AP)^(2)),`along `vec(AP)`
Its axail component is
`dEcostheta=1/(4piepsilon_(0))cdot(lamdadl)/((AP)^(2))x/((AP))=1/(4piepsilon_(0))cdot(lamdaxdl)/((AP)^(3))`
The component of dE perpendicular to the axis = dEsin`theta`.
For charges distributed throughout the entire ring, `sigmadEsintheta = theta`, due to symmetry. Therefore, the resultant axial electric field at the point P,
`E=sumdEcostheta=sum1/(4piepsilon_(0))cdot(lamdaxdl)/((AP)^(3))`
`=1/(4piepsilon_(0))cdotq/(2pir)cdotx/((r^(2)+x^(2))^(3//2))cdotsumdl`
`[becauselamda=q/(2pir)` and `AP=(r^(2)+x^(2))^(1//2)]`
`=1/(4piepsilon_(0))cdotq/(2pir)cdotx/((r^(2)+x^(2))^(3//2))cdot2pir`
`=1/(4piepsilon_(0))cdot(qx)/((r^(2)+x^(2)^(3//2))`