In a certain region of x-y plane, potential function in given in COLUMN-I and corresponding electric lines of  force are given in COLUMN-II. Match the potential function of COLUMN-I with their respective field line representation in COLUMN-II

COLUMN-I (Potential Function)	COLUMN-II (Electric Lines of Force)
A)  $V=x^2-y^2$	p)
B) V = xy	q)
C)  $V=x^2+y^2$	r)
D) $V=\frac{x}{y}$	s)

Since  $E_x=-\frac{\partial v}{\partial x}and{E}_y=-\frac{\partial v}{\partial y}$. Also, tangent to electric lines of force will give direction of  electric field. So,  $\frac{dy}{dx}=\frac{E_y}{E_x}$

A)  $V=x^2-y^2$ $\Rightarrow \overrightarrow{E}=-2x\widehat{i}+2y\widehat{j} \Rightarrow \frac{dy}{dx}=\frac{E_y}{E_x}=\frac{-y}{x}$$\Rightarrow \log y=-\log x+c$

$\Rightarrow xy=c\text{\hspace{0.17em} (Rectangular hyperbola)}$

Corresponding curve is drawn for electric lines of force.
B) V=xy
$\overrightarrow{E}=-y\widehat{i}-x\widehat{j} \Rightarrow \frac{dy}{dx}=\frac{x}{y}\Rightarrow y^2=x^2+c$

Corresponding curve for field lines is shown
C)  $V=x^2+y^2$  $\Rightarrow \overrightarrow{E}=-2x\widehat{i}+2y\widehat{j}$
Tangent to electric line of force will give direction of electric field. So
$\frac{dy}{dx}=\frac{E_y}{E_x}=\frac{y}{x}\Rightarrow \frac{dy}{y}=\frac{dx}{x} \Rightarrow \log y=\log x+c \Rightarrow \frac{y}{x}=c \Rightarrow y=cx$

Straight line equation passing through origin

$$\begin{gathered} D) V=\frac{x}{y} \\ \overrightarrow{E}=-\frac{1}{y}\widehat{i}+\frac{x}{y^2}\widehat{j} \\ \frac{dy}{dx}=\frac{\frac{x}{y^2}}{-\frac{1}{y^2}}=-\frac{x}{y} \end{gathered}$$ 

$\Rightarrow x^2+y^2=Cons\tan t \left(circle\right)$

Corresponding curve is given.