Assertion (A): A charge q is placed at a heig | vedclass

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

Question

(D) The flux through a surface depends on the solid angle subtended by the surface at the position of the charge.For a square of side $b$ with a charg

Vedclass · Accepted Answer

If both $Assertion$ and $Reason$ are false.

Vedclass · Answer

(D) The flux through a surface depends on the solid angle subtended by the surface at the position of the charge.For a square of side $b$ with a charge $q$ at height $d = h/4$ above its center,the solid angle $\Omega$ subtended by the square at the charge is given by $\Omega = 4 \arcsin \left( \frac{b^2}{b^2 + 4d^2} \right)$.The flux $\phi$ is given by $\phi = \frac{q \Omega}{4\pi \epsilon_0}$.Since the solid angle $\Omega$ depends on the ratio of the side length $b$ to the height $d$,the flux depends on the side length $b$. Thus,the $Assertion$ is false.Gauss's law states that the total flux through a closed surface is $q_{enclosed} / \epsilon_0$,which is independent of the shape or size of the Gaussian surface. Thus,the $Reason$ is true.Therefore,the correct option is $D$.

$Assertion (A):$ $A$ charge $q$ is placed at a height $h/4$ above the center of a square of side $b$. The flux associated with the square is independent of the side length $b$.
$Reason (R):$ Gauss's law is independent of the size of the Gaussian surface.

Similar Questions

The figure shows the electric field lines of three charges with charges $+1, +1$,and $-1$. The Gaussian surface in the figure is a sphere containing two of the charges. The total electric flux through the spherical Gaussian surface is

Two charges of $5 Q$ and $-2 Q$ are situated at the points $(3 a, 0)$ and $(-5 a, 0)$ respectively. The electric flux through a sphere of radius $4 a$ having its center at the origin is

$A$ point source of $3000 \ lumen$ is located at the centre of a cube of side length $2 \ m$. The flux through one side is......$lumen$.

Give a reason: 'If the net flux associated with a closed surface is zero,then the net charge enclosed by that surface is zero.'

When is electric flux said to be positive,negative,or zero?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$Assertion (A):$ $A$ charge $q$ is placed at a height $h/4$ above the center of a square of side $b$. The flux associated with the square is independent of the side length $b$.$Reason (R):$ Gauss's law is independent of the size of the Gaussian surface.