A thread carrying a unifrom charge `lambda` per unit length has the configuration shown in fig, Assuming a curvature radius `R` to be considerably less than the length of the thread, find the magnitude fo the electric field strength at the point `O`.
,

Explanation

(a) Using the solution of `3.14`, the net electric field strength at the point `O` due to straight parts of the thread equals is zero. For the curved part (arc) let us derive a general expression i.e., Let us calculate the field strength at the centre of arc fo radius `R` and linear charge density `lambda`and which sutends angle `theta_(0)` at the centre.
From the symmetry the sought field strength will be directed along the bisector of the angle `theta_(0)` and is given by
`E = int_(-theta_(0)//2)^(+theta_(0)//2) (lambda(R d theta)/4pi epsilon_(0) R^(2)) cos theta = (lambda)/(2pi epsilon_(0) R) "sin" (theta_(0))/(2)`
In our problem `theta_(0) = pi//2`, thus the field strength due to the turned part at the point `E_(0) = (sqrt(2) lambda)/(4pi epsilon_(0) R)` which is also the sought result.

(b) Using teh solution fo `3.14` (a) net field strength at `O` due to straight parts equals `sqrt(2) ((sqrt(2) lambda)/(4pi epsilon_(0) R)) = (lambda)/(2pi epsilon_(0) R)` and is directed vertically down, Now using the solving of `3.15`
(a), field strength due to the given curved part (semi-circle) at the point `O` becomes `(lambda)/(2pi epsilon_(0) R)` and is directed vertically upward. Hence the sought net field strength becomes zero.