A rectangular surface of sides 10 ,cm and 15 | vedclass

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

Question

(B) The electric flux $\phi$ through a surface is given by the formula $\phi = E A \cos 	heta$,where $	heta$ is the angle between the electric field

Vedclass · Accepted Answer

$0.1875$

Vedclass · Answer

(B) The electric flux $\phi$ through a surface is given by the formula $\phi = E A \cos 	heta$,where $	heta$ is the angle between the electric field vector $\vec{E}$ and the area vector $\vec{A}$.Given,the surface makes an angle of $30^{\circ}$ with the electric field. The area vector $\vec{A}$ is perpendicular to the surface. Therefore,the angle $	heta$ between the area vector $\vec{A}$ and the electric field $\vec{E}$ is $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.Area $A = 10 \,cm 	imes 15 \,cm = 0.1 \,m 	imes 0.15 \,m = 0.015 \,m^2$.Electric field $E = 25 \,V/m$.Flux $\phi = E A \cos 60^{\circ} = 25 	imes 0.015 	imes 0.5 = 0.1875 \,Nm^2/C$.

$A$ rectangular surface of sides $10 \,cm$ and $15 \,cm$ is placed inside a uniform electric field of $25 \,V/m$,such that the surface makes an angle of $30^{\circ}$ with the direction of the electric field. Find the flux of the electric field through the rectangular surface in $Nm^2/C$.

Similar Questions

$A$ charge $q$ is placed at one corner of a cube as shown in the figure. The flux of the electrostatic field $\overrightarrow{E}$ through the shaded area is ...... .

Draw the electric field lines for a positive point charge.

Select the correct statement regarding electric field lines.

The electric field in a region is uniform and is given by $\vec{E} = a \hat{i} + b \hat{j} + c \hat{k}$. The electric flux associated with a surface of area $\vec{A} = \pi R^2 \hat{i}$ is:

$A$ point charge '$Q$' is placed at a point inside a hollow conducting sphere. Which of the following electric field patterns is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module