Find the electric field intensity at the centre of a semi circular arc of radius r, uniformaly charged with a charge q.

Explanation

In the Fig. 2.30, the length of the semicircular wire is `pir`.
Linear charge density, `lamda= q/(pir)`
Let the charge of a small part dl of the wire be dq.
`thereforedq=lamdacdotdl=q/(pir)cdotdl`
Field intensity at the centre O due to the small part,
`dE=1/(4piepsilon_(0))cdot(dq)/(r^(2))=1/(4piepsilon_(0))cdot(qdl)/(pircdotr^(2))`
`=1/(4piepsilon_(0))cdot(qdtheta)/(pir^(2))[becausedl=rdtheta]`
`

Now the field intensity dE is resolved into two components, one along the radius which is `dE_(x) = dEcostheta` and another perpendicular to the radius which is `dE_(y) = dEsintheta`. Considering the whole wire, it is seen that all the `dE_(x)` components get cancelled. Only the `dE_(y)` components get remain.
`therefore` Total field intensity at O,
`E=int_(0)^(pi)dEsintheta=1/(4piepsilon_(0))cdotq/(pir^(2))int_(0)^(pi)sintheta.dtheta`
`=1/(4piin_(0))cdotq/(pir^(2))(1-cospi)=1/(4piin_(0))cdot(2q)/(pir^(2))`