You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introducing Flux

Consider an e...

Question

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

2 E_{0} a^{2}

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

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

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

E_{0} a^{2}

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

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

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

Open in App

Solution

The correct option is D $\frac{E_{0} a^{2}}{\sqrt{2}}$
on locating the plate is tilled lly $45^{o}$
$ϕ = E . S$
$= E_{0} a^{2} cos 45$
$= \frac{E_{0} a^{2}}{\sqrt{2}}$

Suggest Corrections

Similar questions

¯ ¯¯ ¯ E = E_{0}^x

E_{0}

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

E_{0} = 2 \times 10^{3} N C^{- 1}

0.2 m^{2}

Y Z

E_{0}

\vec{E} = \frac{3}{5} E_{0} \vec{i} + \frac{4}{5} E_{0} \vec{j} with E_{0} = 2 \cdot 0 \times 10^{3} N C^{- 1}

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

E_{0} = 2 \times 10^{3} N C^{- 1}

0.2 m^{2}

Y - Z

Join BYJU'S Learning Program