Conisder an eletric field $\vec{E} =E_0\hat{x},~where~E_0$ is a constant . The flux through the shaded area (as shown in the figure) due to this field is

E_{0} a^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{2} E_{0} a^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{E_{0} a^{2}}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 E_{0} a^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $E_{0} a^{2}$
$\to A = \frac{1}{2} \times {\to d}_{1} \times {\to d}_{2}$

$= \frac{1}{2} (a^i + a^j + a^k) \times (a^i - a^j + a^k)$

$ϕ = \to E . \to A {\to E i s c o n s t a n t}$

$= E_{0} a^{2}$

Suggest Corrections

Similar questions

Q. Consider an electric field

¯ E = E_{0}^x

where

E_{0}

is a constant. The flux through the shaded area (as shown in the figure) due to this field is

Q. Consider an electric field

¯ ¯¯ ¯ E = E_{0}^x

, where

E_{0}

is a constant. The flux through the shaded area (as shown in the figure) due to this field is:

Q. The electric field in a region of space is given by

\to E = 5 ˆ i + 2 ˆ j N / C

. The flux of

\to E

due to this field through an area

1 m^{2}

lying in the y-z plane, in SI units, is :

The electric field in a region of space is given by $\to E = (^5 i +^2 j) N c^{- 1}$ . The electric flux due to this field through an area $2 m^{2}$ lying in the Y-Z plane in S.I. units is

Q. The electric field in a region is given by

\to E = \frac{3}{5} E_{0} \to i + \frac{4}{5} E_{0}^j

with

E_{0} = 2.0 \times 10^{3} N C^{- 1}

. Find the flux of this field through a rectangular surface of area

0.2 m^{2}

parallel to the

y - z

plane.

Join BYJU'S Learning Program

Conisder an eletric field \(\vec{E} =E_0\hat{x},~where~E_0\) is a constant . The flux through the shaded area (as shown in the figure) due to this field is

The correct option is A E0a2→A=12×→d1×→d2 =12(a^i+a^j+a^k)×(a^i−a^j+a^k) ϕ=→E.→A{→E is constant} =E0a2

The correct option is A $E_{0} a^{2}$
$\to A = \frac{1}{2} \times {\to d}_{1} \times {\to d}_{2}$

$= \frac{1}{2} (a^i + a^j + a^k) \times (a^i - a^j + a^k)$

$ϕ = \to E . \to A {\to E i s c o n s t a n t}$

$= E_{0} a^{2}$