The electric field in a region is given by E¯=25E0i^+35E0j^ with E0=4.0×103N/c . The flux of this field through a rectangular surface area 0.4 m2 parallel to the y-z plane is (in Nm2/c)

ϕ=E¯ . A¯=(25E0i^+35E0j^).(0.4i^)=25E0(0.4)=25(4×103)(0.4)=640 NM2/C

The electric field in a region is given by E¯=25E0i^+35E0j^ with E0=4.0×103N/c . The flux of this field through a rectangular surface area 0.4 m2 parallel to the y-z plane is (in Nm2/c)

ϕ=E¯ . A¯=(25E0i^+35E0j^).(0.4i^)=25E0(0.4)=25(4×103)(0.4)=640 NM2/C

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The electric field in a region is given by $\bar{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} N / c$ . The flux of this field through a rectangular surface area 0.4 $m^{2}$ parallel to the y-z plane is (in N $m^{2}$ /c)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

960

1600

640

800

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$ϕ = \bar{E} . \bar{A} = (\frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}) . (0.4 \hat{i}) = \frac{2}{5} E_{0} (0.4) = \frac{2}{5} (4 \times 10^{3}) (0.4) = 640 N M^{2} / C$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

The electric field in a region is given by E¯=25E0i^+35E0j^ with E0=4.0×103N/c . The flux of this field through a rectangular surface area 0.4 m2 parallel to the y-z plane is (in Nm2/c)

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App

The electric field in a region is given by $\bar{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} N / c$ . The flux of this field through a rectangular surface area 0.4 $m^{2}$ parallel to the y-z plane is (in N $m^{2}$ /c)