Question

Under certain conditions the interaction between a "polar" molecule such as$\mathit{H}\mathit{C}\mathit{l}$located at the origin and an ion located along the x axis can be described by a potential energy $\mathit{U}\mathbf{=}\mathbf{-}\mathit{b}\mathbf{/}{\mathit{x}}^{\mathbf{2}}$, where b is a constant. What is ${\mathit{F}}_{\mathbf{x}}$, the $\mathbf{x}$component of the force on the ion? What is ${\mathit{F}}_{\mathbf{y}}$, the $\mathbf{y}$component of the force on the ion?

Short answer

$F_x$, the $x$ component of the force on the ion is $-2bx$

$F_y$, the $y$ component of the force on the ion is 0

Step-by-step solution

Identification of given data

Potential energy$\mathit{U}\mathbf{=}\mathbf{-}\mathit{b}\mathbf{/}{\mathit{x}}^{\mathbf{2}}$

concept of the potential energy

Potential energy is calculated when the energy kept in an object due to the position of an object is related to zero.

Determination of Fx, thex component of the force and Fy, the y component of the force

The$x$component of the force,

$F_x=-\frac{\partial U}{\partial x}$

$=-\frac{\partial }{\partial x}\left(\frac{-b}{x^2}\right)$

$=b\frac{\overrightarrow{\partial }}{\partial x}\left(x^{-2}\right)$

$=b(-2x)$

$=-2bx$

Hence, the x component of the force on the ion is $-2bx$

The y component of the force,

$F_y=-\frac{\partial U}{\partial y}$

$=-\frac{\partial }{\partial y}\left(\frac{-b}{x^2}\right)$

$=b\frac{\partial }{\partial y}\left(x^{-2}\right)$

$=0$

Hence, the $y$component of the force on the ion is $0$.