A long wire with a uniform charge density `lambda` is bent in two configurations shown in fig. Determine the electric field intensity at point O.

Explanation

For Fig. Field due to segment 1 is
`vecE_(1) =[(lambda )/(4 pi epsilon__(0)R)]hat i+[-(lambda )/(4 pi epsilon__(0)R)]hat j`
field dur to segment 2 is
`vecE_(2) =[(lambda)/(4 pi epsilon_(0)R)]hat i+[-(lambda )/(4 pi epsilon_(0)R)]hat j`
Field due tp quater shape wire segment 3 is
`vecE_(3) =[(lambda )/(4 pi epsilon__(0)R)]hat i+[-(lambda )/(4 pi epsilon__(0)R)]hat j ( :. theta_(1) =90^(@), theta_(2)=0^(@))`
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The resultant field is the superpositon of the fields due to each part, i.e.
`vec(E)=vecE_(1)+vecE_(2)+vecE_(3)` ...(i)
Substituting the values of `vecE_(1), vecE_(2)`, and `vecE_(3 )` in eq. (i) we get,
`vec(E) =[(lambda )/(4 pi epsilon_(0)R)]hat i+[(lambda)/(4 pi epsilon_(0)R)]hat j`
`|vecE_(1)| =[((lambda )/(4 pi epsilon_(0)R))^(2)]+((lambda )/(4 pi epsilon_(0)R))^(2)]^(1//2) =(sqrt(2)lambda)/(4 pi epsilon_(0)R)`
here, `E_(x)=E_(y)=(lambda )/(4 pi epsilon_(0)R)`
Hence, the resulant field will make an angle of `45^(@)` with the axis.
(b). For fig. field due to segment 1 is
`vecE_(1)=(lambda )/(4 pi epsilon_(0)R)[hat i-hatj]`
Field due to segment 2 is
`vecE_(x_(2))=(lambda )/(4 pi epsilon_(0)R)hat i`
`vecE_(y_(2))=(lambda )/(4 pi epsilon_(0)R)hat i`
`vecE_(2 )=(lambda )/(4 pi epsilon_(0)R)[hat i-hatj]`
Field due to segment 3 is
`vecE_(3)=(lambda )/(2 pi epsilon_(0)R)hat j`
from the principle of superposition of electric fields.
`vec(E)=vecE_(1)+vecE_(2)+vecE_(3)`
`=(lambda )/(4 pi epsilon_(0)R)[hat i-hatj]-(lambda )/(4 pi epsilon_(0)R)[hat i-hatj]+(lambda )/(2 pi epsilon_(0)R)[hat i-hatj]=0`
Hence, the net field is zero.