Consider an electric field vec{E} = E_0 hat{i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The shaded area is a square in the $xz$-plane tilted at an angle of $45^{\circ}$ to the $x$-axis. The vertices of the square are $(0,0,0)$,$(0,a,0

Vedclass · Accepted Answer

$E_0 a^2$

Vedclass · Answer

(C) The shaded area is a square in the $xz$-plane tilted at an angle of $45^{\circ}$ to the $x$-axis. The vertices of the square are $(0,0,0)$,$(0,a,0)$,$(a,a,a)$,and $(a,0,a)$.The area vector $\vec{A}$ has a magnitude equal to the area of the square,which is $a 	imes a \sqrt{2} = \sqrt{2} a^2$. The direction of the area vector is perpendicular to the surface. Since the surface lies in a plane defined by $y=z$ (or $z-y=0$),the normal vector is proportional to $\hat{j} - \hat{k}$.Alternatively,we can define the area vector by the cross product of two adjacent sides: $\vec{A} = \vec{AB} 	imes \vec{AD}$. Let $\vec{AB} = a\hat{j}$ and $\vec{AD} = a\hat{i} + a\hat{k}$.Then $\vec{A} = (a\hat{j}) 	imes (a\hat{i} + a\hat{k}) = a^2(\hat{j} 	imes \hat{i}) + a^2(\hat{j} 	imes \hat{k}) = -a^2\hat{k} + a^2\hat{i} = a^2\hat{i} - a^2\hat{k}$.The electric flux $\phi$ is given by $\phi = \vec{E} \cdot \vec{A}$.Given $\vec{E} = E_0 \hat{i}$,we have:$\phi = (E_0 \hat{i}) \cdot (a^2 \hat{i} - a^2 \hat{k})$$\phi = E_0 a^2 (\hat{i} \cdot \hat{i}) - E_0 a^2 (\hat{i} \cdot \hat{k})$Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{k} = 0$,we get:$\phi = E_0 a^2$.

Consider an electric field $\vec{E} = E_0 \hat{i}$,where $E_0$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

Similar Questions

For a closed surface $\oint \vec{E} \cdot d\vec{s} = 0$,then:

An electric dipole is placed in the north-south direction inside a sphere filled with water. Which statement is correct?

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):

$A$ charged particle $q$ is placed at the centre $O$ of a cube of side length $L$ $(A, B, C, D, E, F, G, H)$. Another identical charge $q$ is placed at a distance $L$ from $O$, outside the cube, as shown in the figure. The electric flux through the face $BGFC$ is

$A$ uniformly charged conducting sphere of diameter $14 \ cm$ has a surface charge density of $40 \ \mu C/m^2$. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/N \cdot m^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module