Consider a uniform electric fieldE= 3 × 103îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to theyzplane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with thex-axis?

Dear Sri

Flux φ=E.A

in first case E is along x axis and area vector is also along x axis since the area is parallel to yz plane

φ=EA*cos0⁰=EA=3x10³ x(0.1)² = 30 Nm²/C

in second case

E is along x axis and area vector that is normal is making 60 with x axis

Φ=EA*cos60⁰=EA/2=15 Nm²/C

All the best.                                                           

AKASH GOYAL

AskiitiansExpert-IIT Delhi

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively :Click here to download the toolbar..

Approved

Dear Student,
Please find below the solution to your problem.

Area A=(0.1)2=0.01m2
Flux ϕ=EAcosθ
(i)θ=0o
ϕ=3×103×0.01×cos0o
⟹ϕ=30Nm2/C
(i)θ=60o
ϕ=3×103×0.01×21​
⟹ϕ=15Nm2/C

Thanks and Regards