Two semicircle wires ABC, and ADC, each of radius R are lying on xy and xz planes, respectively as shown in fig.
if the linear charge density of the semicircle parts and straight parts and straight parts is `lambda`, Find the electric field entensity `vec(E)` at the origin .

Answer

`-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`.

Explanation

Electric field due to MA: `vecE_(1)=(lambda)/(4 pi epsilon_(0)R)(hati -hatk)`
Electric field due to ADC: `vecE_(2)=(lambda)/(2 pi epsilon_(0)R)(-hatk)`
Electric field due to ABC: `vecE_(3)=(lambda)/(2 pi epsilon_(0)R)(-hatj)`
Electric field due to NC: `vecE_(4)=(lambda)/(4 pi epsilon_(0)R)(hati-hatk)`
Net electric field is
`vecE_(0)=vecE_(1)+vecE_(2)+vecE_(3)+vecE_(4)=-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`.