Question

In a space, potential $\left(V\right)$ is given by $V={\log }_e\left(xy\right)$, where $V,x,y$ are in SI units. The angle between $\overrightarrow{E}$ & $X$-axis at point $(6,6\sqrt{3})\text{m}$

• A. ${30}^{\circ }$
• B. ${150}^{\circ }$
• C. ${210}^{\circ }$
• D. ${240}^{\circ }$

Answer 1

The correct option is C ${210}^{\circ }$

Electric field is defined as $E=\frac{-dV}{dr}$
Electric field along x-axis
$E_x=\frac{-d}{dx}\{{\log }_{e}x+{\log }_{e}y\}=\frac{-1}{x}$
Electric field along y-axis
$E_y=\frac{-d}{dy}\{{\log }_{e}x+{\log }_{e}y\}=-\frac{1}{y}$

$(E_x,E_y)=(-\frac{1}{x},-\frac{1}{y})$



As $(x,y)$ is $(6,6\sqrt{3})$
$\tan \theta =\frac{E_y}{E_x}=\frac{-\frac{1}{y}}{\frac{-1}{x}}=\frac{x}{y}=\frac{6}{6\sqrt{3}}=\frac{1}{\sqrt{3}}$

As $\theta$ is in third quadrant
Angle between $\overrightarrow{E}$ and $X$ axis is
$=180+30={210}^o$