You visited us 90 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Surface Integral

A plane area ...

Question

A plane area of $100 {c m}^{2}$ is placed in uniform electric field of $100 N / C$ such that the angle between area vector and electric field is $60^{o}$ . The electric flux over the surface is

0.5 Wb

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

5 Wb

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

1 Wb

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

Open in App

Solution

The correct option is A 0.5 Wb
Electric Flux $(ϕ) = \to E . \to A = E A cos θ$
$\Rightarrow ϕ = 0.01 m^{2} \times 100 N / C \times cos 60^{0}$
$\Rightarrow ϕ = 0.5 W b$

Suggest Corrections

Similar questions

Q. A plane surface of area

10 c m^{2}

is placed in a uniform electric field of

20 N / C

such that the angle between the surface and the electric field is

30^{o}

. The electric flux over the surface is:

Q. A circular plane sheet of radius

10 c m

is placed in a uniform electric field of

5 {\times 10}^{5} N C^{- 1}

, making an angle of

60^{o}

with field. The electric flux through the sheet is

A particle that carries a charge $- q$ is placed at rest in uniform electric field $10 N / C$ . It experiences a force and moves in a certain time t, it is observed to acquire a velocity $10 \to i - 10 \to j$ m/s. The given electric field intersects a surface of area $A$ $m^{2}$ in the X -Z plane. Electric flux through surface is:

Q. If the surface normal vector

^n

makes an angle

θ

with the electric field

\to E

, then the electric flux through a surface of area

d S

is given by

Q. A cylinder of radius r and length l is placed in a uniform electric field of intensity E acting parallel to the axis of the cylinder. The total flux over curved surface area is:

Join BYJU'S Learning Program

A plane area of 100 cm2 is placed in uniform electric field of 100 N/C such that the angle between area vector and electric field is 60o. The electric flux over the surface is

The correct option is A 0.5 WbElectric Flux (ϕ)=→E.→A=EAcosθ⇒ϕ=0.01m2×100N/C×cos600⇒ϕ=0.5Wb

A plane area of $100 {c m}^{2}$ is placed in uniform electric field of $100 N / C$ such that the angle between area vector and electric field is $60^{o}$ . The electric flux over the surface is

The correct option is A 0.5 Wb
Electric Flux $(ϕ) = \to E . \to A = E A cos θ$
$\Rightarrow ϕ = 0.01 m^{2} \times 100 N / C \times cos 60^{0}$
$\Rightarrow ϕ = 0.5 W b$