Given a uniform electric field E = 5 × 103 N / C , find the flux of this field through a square of side 10 cm on a side whose plane is parallel to y -z plane. What would be the flux through the same square. If the plane makes an angle of 30 with the x - axis.
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Answer:magnitude of electric field intensity=5×10*3 NC*-1
SIDE OF SQ.=10cm=0.1m
Area of sq.=0.1×0.1=0.01m*2
The plane of sq. Is parallel
So, angle=0
Flux=E×A COS0
=5×10*3×0.01
=50Nm*2C*-1
If plane makes angle of 30 degree with x-axis then theta=60 degree
Flux=5×10*3×0.01×cos60degree
=25Nm*2C-1
Explanation:
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