A rod of length L has a total charge Q distributed uniformly along its length.It is bent in the shape of a semicircle.Find the magnitude of the electric field at the centre of curvature of the semicircle.

Here, since length of rod is L, So we can write:Perimeter of semi-circular arc = Length of rodπR=L, where R is radius of curvatureNow we know that, For a circular arc subtending angle ¢(phi) at the centre, the magnitude of electric field at centre of curvature is given by:E=[2kQsin(¢/2)]/¢R^2 , where R is radius for curvature.So here, in the given condition:¢=π, R=L/π ( from the equation of first step)So now electric field at centre of curvature in given condition will be:E=[2kQsin(π/2)]/π(L/π)^2E=2πkQ/L^2Hope it helpsThanku very much