JEE Main 2014 · 19 Apr · Shift 1 · Physics Q16

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 16 of 90

The electric field in a region of space is given by, $\vec E=E_o\hat i+2E_o\hat j$ where $E_0=100$ N/C. The flux of this field through a circular surface of radius 0.02 m parallel to the Y-Z plane is nearly :

0.125 Nm

^2

0.02 Nm

^2

0.005 Nm

^2

3.14 Nm

^2

Answer and worked solution

Official answer: A

A disc parallel to the Y-Z plane has its normal along $x$ , so only $E_x=E_0$ contributes:
$\Phi=E_0\pi r^2=100\times\pi\times(0.02)^2\approx0.125\text{ N m}^2\text{/C}.$

Topics:Electric flux and Gauss's law

← Previous question Next question →

JEE Main 2014 · 19 Apr · Shift 1, Physics question 16