consider a uniform electric field e is equal to 300 Newton per coulomb what should be the electric flux through the square of 10 cm on a side whose plane is parallel to y z plane what is the flux through the same square if the normal to its plane makes an angle of 60 degree with x axis
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Answer:
Electric field intensity,
E
=3×10
3
N/C
Magnitude of electric field intensity, ∣E∣=3×10
3
N/C
Side of the square, s=10cm=0.1m
Area of the square, A=s
2
=0.01m
2
The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ=0
o
Flux (ϕ) through the plane is given by the relation,
ϕ=∣
E
∣Acosθ
=3×10
3
×0.01×cos0
o
=30Nm
2
/C
(b)
Plane makes an angle of 60
o
with the x-axis. Hence, θ=60
o
Flux, ϕ=∣
E
∣Acosθ
=3×10
3
×0.01×cos60
o
=30×
2
1
=15Nm
2
/C
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