$Assertion (A):$ $A$ 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.

  • A
    If both $Assertion$ and $Reason$ are true and $Reason$ is the correct explanation of $Assertion$.
  • B
    If both $Assertion$ and $Reason$ are true but $Reason$ is not the correct explanation of $Assertion$.
  • C
    If $Assertion$ is true but $Reason$ is false.
  • D
    If both $Assertion$ and $Reason$ are false.

Explore More

Similar Questions

$A$ charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $P$, as shown in the figure. When $\theta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$. If $\theta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is. . . . . . .

$A$ $6 \mu C$ charge is placed at the centre of a cube. What will be the electric flux through each face of the cube? (Take $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

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

When does the electric flux associated with a closed surface become positive,zero,or negative?

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo