wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The cube as shown in Fig. has sides of length L=10.0 cm. The electric field is uniform, has a magnitude E=4.00×103NC1, and is parallel to the xyplane at an angle of 37 measured from the +xaxis towards the +yaxis.
The total net electric flux through all faces of the cube is
160283_75c8b1dca37d4e0bb3fe100a7114d40b.png

A
8Nm2C1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8Nm2C1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
24Nm2C1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Zero
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D Zero
Here, the electric field is given by E=Ecos37^i+Esin37^j
Electric flux through any surface S is ϕ=E.^nS
Where ^n is the normal outwardto the surface S
For surface S1: ^n=^j,S1=L2=(0.1)2m2
Thus, ϕ1=E.^nS1=(Ecos37^i+Esin37^j).(L2^j)=EL2sin37=(4×103)(0.1)2sin37=24Nm2C1
For surface S2: ^n=^k,S2=L2=(0.1)2m2
Thus, ϕ2=E.^nS2=(Ecos37^i+Esin37^j).(L2^k)=0
For surface S3: ^n=^j,S3=L2=(0.1)2m2
Thus, ϕ3=E.^nS3=(Ecos37^i+Esin37^j).(L2^j)=EL2sin37=(4×103)(0.1)2sin37=24Nm2C1
For surface S4: ^n=^k,S4=L2=(0.1)2m2
Thus, ϕ4=E.^nS4=(Ecos37^i+Esin37^j).(L2^k)=0
For surface S5: ^n=^i,S5=L2=(0.1)2m2
Thus, ϕ5=E.^nS5=(Ecos37^i+Esin37^j).(L2^i)=EL2cos37=(4×103)(0.1)2cos37=32Nm2C1
For surface S6: ^n=^i,S6=L2=(0.1)2m2
Thus, ϕ6=E.^nS6=(Ecos37^i+Esin37^j).(L2^i)=EL2cos37=(4×103)(0.1)2cos37=32Nm2C1
Thus, net flux =ϕ1+ϕ2+ϕ3+ϕ4+ϕ5+ϕ6=24+0+24+0+3232=0

580693_160283_ans_d18d42b8914940ebbc29db5256600762.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Problem Solving Using Newton's Laws
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon