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:

Complete Solution:

The force on (q) is the vector sum of the forces due to the two charges (Q).

The force due to one charge (Q) on (q) is given by Coulomb’s law:

F = kQq / r²,

where (r) is the distance between (Q) and (q):

r = √((d/2)² + x²)

So,

F = kQq / ((d² / 4) + x²)

Net Force:

The horizontal components of the forces cancel out, leaving only the vertical components. The net force on \(q\) is:

F_net = 2F sinθ,

where:

sinθ = x / r = x / √((d/2)² + x²)

Substituting values:

F_net = (2kQq * x) / ((d² / 4) + x²)³/²

To maximize (F_net), differentiate it with respect to (x) and set the derivative to 0:

f(x) = x / ((d² / 4) + x²)³/²

After differentiation and simplification:

f'(x) = [((d² / 4) + x²) - 3x²] / ((d² / 4) + x²)⁵/² = 0

Solving:

(d² / 4) - 2x² = 0

x² = d² / 8

Taking the square root:

x = d / 2√2

Final Answer:

The value of (x) at which the charge (q) experiences the maximum Coulomb force is:

x = d / 2√2