Step 1:

Length of the rod is .

Charge on each glass rod is .

Distance between two glass rods is .

Electric field strengths   and are placed at and to the right of left side rod by connecting mib points of two rods.

Diagram of the situation:

For a non conducting rod of length with a uniform charge density and a total charge is lying along axis.

Electric field strength at a point located at a distance from  the center of the rod along its perpendicular bisector is .

Field strength on each position is resultant of both left and right rods.

(i)

Find Electric field strength at the position  .

Elcetric field strength Contribution of left rod:

Diastance from the left rod to  the position  is .

Substitute corresponding values in the formula.

.

Elcetric field strength Contribution of right rod:

Diastance from the left rod to  the position  is .

Substitute corresponding values in the formula.

.

Direction of field strength of right rod is in opposite to that of field strength of left rod.

Thus, .

Electric field strength at the position  is .

(ii)

Find Electric field strength at the position  .

is exactly midpoint between the two rods.

Since moid point is considered , the magntude of the right rod contribution is also same as the left rod and is in opposite direction.

.

(iii)

Find Electric field strength at the position  .

Elcetric field strength Contribution of left rod:

Diastance from the left rod to  the position  is .

Substitute corresponding values in the formula.

.

Elcetric field strength Contribution of right rod:

Diastance from the left rod to  the position  is .

Substitute corresponding values in the formula.

.

Direction of field strength of right rod is in opposite to that of field strength of left rod.

Thus, .

Electric field strength at the position  is .