(D) The electric field is given as $\vec{E} = 1 \hat{j} \ N/C$.Since the square is placed in the $XY$-plane,its area vector $\vec{A}$ is perpendicular

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(D) The electric field is given as $\vec{E} = 1 \hat{j} \ N/C$.Since the square is placed in the $XY$-plane,its area vector $\vec{A}$ is perpendicular to the $XY$-plane,which is along the $Z$-axis.Therefore,$\vec{A} = 1 \hat{k} \ m^2$.The electric flux $\phi$ is defined as the dot product of the electric field and the area vector:$\phi = \vec{E} \cdot \vec{A}$$\phi = (1 \hat{j}) \cdot (1 \hat{k})$Since the dot product of orthogonal unit vectors $\hat{j} \cdot \hat{k} = 0$,the flux is:$\phi = 0 \ Nm^2/C$.

Class 12 Physics Electric Charges and Fields Electric Field Lines, Electric Flux and Gauss's Law

Easy

There exists an electric field of $1 \ N/C$ along the $Y$-direction. The flux passing through a square of $1 \ m^2$ placed in the $XY$-plane inside the electric field is . . . . . . .

GUJCET 2011 All GUJCET

A
$1.0 \ Nm^2/C$
B
$10.0 \ Nm^2/C$
C
$2.0 \ Nm^2/C$
D
zero

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Class 12

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Chapter

Electric Charges and Fields

Subtopic

Electric Field Lines, Electric Flux and Gauss's Law

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GUJCET 2011 Paper All GUJCET years →

The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.

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$A$ point charge $+10 \; \mu C$ is at a distance $5 \; cm$ directly above the centre of a square of side $10 \; cm$,as shown in the figure. What is the magnitude of the electric flux through the square?

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If the number of electric lines of force emerging out of a closed surface is $1000$,then the charge enclosed by the surface is .......... $C$.

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In a region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

EasyAP EAMCET 2025

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$A$ spherical rubber balloon carries a charge $q$ uniformly distributed over its surface. As the balloon is inflated,the total electric flux coming out of the surface:

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There exists an electric field of $1 \ N/C$ along the $Y$-direction. The flux passing through a square of $1 \ m^2$ placed in the $XY$-plane inside the electric field is . . . . . . .

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The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.

$A$ point charge $+10 \; \mu C$ is at a distance $5 \; cm$ directly above the centre of a square of side $10 \; cm$,as shown in the figure. What is the magnitude of the electric flux through the square?

If the number of electric lines of force emerging out of a closed surface is $1000$,then the charge enclosed by the surface is .......... $C$.

In a region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

$A$ spherical rubber balloon carries a charge $q$ uniformly distributed over its surface. As the balloon is inflated,the total electric flux coming out of the surface:

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