Question

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 Coulombs force is :

• A. x = d
• B. $x = \frac{d}{2}$
• C. $x = \frac{d}{\sqrt2}$
• D. $x = \frac{d}{2\sqrt2}$

Answer 1

Answer: (D)

$x = \frac{d}{2\sqrt2}$

Answer 2

The correct answer is (D) : $x = \frac{d}{2\sqrt2}$

Fig.

$F = \frac{KQq}{[x^2 + \frac{d^2}{4}]}$
Net force on g = 2 F cos$\theta$
$F_{net} = \frac{2KQqx}{[x^2+\frac{d^2}{4}]^{3/2}}$
For maximum $F_{net}$
$\frac{dF_{net}}{dx} = 0$
$$we get x = $\frac{d}{2\sqrt2}$