Two point charges A and B of magnitude + 8 × 10 ^ - 6 C and - 8 ×...

Direction of electric field at mid-point due to both charge will be in same direction, which is towards right.

Therefore, $E$ at mid-point $= \dfrac{kq}{{\left(\dfrac{d}{2}\right)}^{2}}$ + \dfrac{kq}{{\left(\dfrac{d}{2}\right)}^{2}}

$⇒ 6 . 4 \times {10}^{4} = 2 \times \left[9 \times {10}^{9} \times \dfrac{\left(8 \times {10}^{- 6}\right)}{{d}^{2}} \times 4\right]$

$⇒ d = 3m$