The electric field intensity at all points in space is given by E → = 3 i ^ - 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 be

Question

The electric field intensity at all points in space is given by $\vec{E} = \sqrt{3} \hat{i} - \hat{j}$ 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 $θ = 6 0^{\circ}$ then the flux of electric field through frame is

TestHub · Accepted Answer

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

The magnitude of the electric field is $E = \sqrt{E_{x}^{2} + E_{y}^{2}} = \sqrt{3 + 1} = 2 {V m}^{- 1}$

The direction of the electric field is given by $θ = {tan}^{- 1} \frac{E_{y}}{E_{x}} = {tan}^{- 1} \frac{1}{\sqrt{3}} = 3 0^{\circ}$

Hence electric field is normal to square frame $LMNO$ as shown in the figure.

$∴$ Electric flux $= \vec{E} \cdot \vec{A} = E A cos θ = 2 \times 1 = 2 V m$

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