Question

The electric field intensity at all points in space is given by $\overset{ \rightarrow }{E}=\sqrt{3}\hat{i}-\hat{\text{j}}$ volts/metre. A square frame $LMNO$ of side 1 metre is shown in the figure. The point $N$ lies in $x-y$ plane. The angle between the line $ON$ and the $x$ -axis is $\theta = 6 0^{^\circ }$ then the flux of electric field through frame is

• A. $0Vm$
• B. $1 \, V \, m$
• C. $2 \, V \, m$
• D. $4 \, V \, m$

Answer 1

Answer: (C)

The direction of the electric field is in $x-y$ plane as shown in figure

The magnitude of the electric field is $\textit{E}=\sqrt{\textit{E}_{\text{x}}^{2} + \textit{E}_{\text{y}}^{2}}=\sqrt{3 + 1}=2 \, \text{V m}^{- 1}$
The direction of the electric field is given by $\theta =\text{tan}^{- 1}\frac{\textit{E}_{\text{y}}}{\textit{E}_{\text{x}}}=\text{tan}^{- 1}\frac{1}{\sqrt{3}}=30^{^\circ }$
Hence electric field is normal to square frame $LMNO$ as shown in the figure.
$\therefore $ Electric flux $=\overset{ \rightarrow }{E}\cdot \overset{ \rightarrow }{A}=\textit{E A}\text{ cos}\theta =2\times 1=2 \, V \, m$