Question

The electric field intensity at all points in space is given by $\overrightarrow{E}=\sqrt{3}\hat{i}-\hat{\text{j}}$ 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 $\theta =60^{{}^{\circ }}$ then the flux of electric field through frame is

• A. $0Vm$
• B. $1Vm$
• C. $2Vm$
• D. $4Vm$

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