The figure shows some of the electric field lines corresponding to an electric field. The figure suggests

  • A
    ${E_A} > {E_B} > {E_C}$
  • B
    ${E_A} = {E_B} = {E_C}$
  • C
    ${E_A} = {E_C} > {E_B}$
  • D
    ${E_A} = {E_C} < {E_B}$

Explore More

Similar Questions

$A$ charge $q$ is placed at the corner of a cube of side $a$. The electric flux through the cube is

$A$ 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 $Nm^2/C$.

$A$ charge $q$ is placed at the center of the circular base of an inverted cone of height $h$ and base radius $R$. The cone is capped by a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in $SI$ units). The value of $n$ is. . . .

$A$ square loop of sides $a=1 \ m$ is held normally in front of a point charge $q=1 \ C$. The charge is placed at a distance of $a/2$ from the center of the square. The flux of the electric field through the shaded region is $\frac{5}{p} \times \frac{1}{\varepsilon_0} \frac{N m^2}{C}$,where the value of $p$ is . . . . . . .

$A$ charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface 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