Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

Step-by-step solution

The distance between q and Q is r = x 2 + d 2 2 = x 2 + d 2 4 . The force between q and Q will be F = k q Q r 2 . From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore, F n e t = 2 k q Q r 2 cos θ = 2 k q Q r 2 × x r = k q Q x x 2 + d 2 4 3 2 For force to be maximum, d F n e t d x = 0 . ⇒ k q Q x 2 + d 2 4 3 2 - x × 3 2 x 2 + d 2 4 1 2 × 2 x x 2 + d 2 4 3 2 2 = 0 ⇒ x 2 + d 2 4 = 3 x 2 &#8