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Consider a uniform electric field E = 3 × 10 3 îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x -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..
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..
Dear Student,Please find below the solution to your problem.Area A=(0.1)2=0.01m2Flux ϕ=EAcosθ(i)θ=0oϕ=3×103×0.01×cos0o⟹ϕ=30Nm2/C(i)θ=60oϕ=3×103×0.01×21⟹ϕ=15Nm2/CThanks and Regards
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