Question

Four equal charges $+q$ are placed at four corners of a square with its center as origin and lying in $\text{yz}$ plane. The electrostatic potential energy of a fifth charge $+q$ varies on $\text{x-}$axis as

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Here $x$ will be the axial location for the square. If we move along $x-$axis from center for any charge the distance $r$ of the point will be


$r^2=x^2+\frac{a^2}{2}$

$\Rightarrow r=\sqrt{x^2+\frac{a^2}{4}}$

Now potential energy of charge $+q$ at any distance $x$ from origin along $\text{x-}$axis is given,

$U_P=\frac{4k{q}^2}{r}$

$\Rightarrow U_P=\frac{4k{q}^2}{\sqrt{x^2+\frac{a^2}{4}}}$

Differentiating the above equation w.r.t $x$

$\frac{d{U}_p}{dx}=-\frac{8k{q}^{2}x}{2{(x^2+\frac{a^2}{4})}^{3/2}}$​

Now, as we can see for $x<0$:

The slope potential energy curve, $\frac{d{U}_p}{dx}>0$, means the slope is positive.

For $x=0$:

$\frac{d{U}_p}{dx}=0$, means the slope is zero.

For $x>0$:

$\frac{d{U}_p}{dx}<0$ means, the slope is negative.

Hence, option (b) is the correct alternative.