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

$A$ point charge is distributed in a small volume. The electric flux through a spherical surface of radius $10 \ cm$ enclosing the charge is $20 \ Vm$. What is the electric flux through a concentric spherical surface of radius $20 \ cm$?

Gauss's law is true only if the force due to a charge varies as:

If a charge $Q$ is placed at the corner of a cube,what is the electric flux passing through one of its faces?

$A$ cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

The figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. The total flux of the electric field through a closed surface $S$,due to all charges $q_1, q_2, q_3$ and $q_4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module