JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Electrostatics

The electric field in a region is given by

E = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \frac{N}{C}

. The ratio of flux of reported field through the rectangular surface of area

0.2 m^{2}

(parallel to

y - z

plane) to that of the surface of area

0.3 m^{2}

(parallel to

x - z

plane) is

a : b

, where

a =

____
[Here

\hat{i}, \hat{j}

and

\hat{k}

are unit vectors along

x, y

and

z -

axes respectively]

Answer: 1

Solution

Flux counts only the field component along the normal to the surface. A rectangle parallel to the

y

z

plane has its normal along

\hat{i}

, so only the

\hat{i}

part of the field crosses it:

ϕ_{1} = \frac{3}{5} E_{0} \times 0.2 = 0.12 E_{0}

A rectangle parallel to the

x

z

plane has its normal along

\hat{j}

, so only the

\hat{j}

part crosses it:

ϕ_{2} = \frac{4}{5} E_{0} \times 0.3 = 0.24 E_{0}

Their ratio is

0.12 : 0.24 = 1 : 2

, so

a = 1

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Integer answer