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प्रश्न
The electric field in a region is given by
`vec"E"= 3/5"E"_0 vec"i" + 4/5 "E"_0 vec "i" "with" " E"_0 = 2.0 xx 10^3 "N""C"^-1.`
Find the flux of this field through a rectangular surface of area 0⋅2 m2 parallel to the y-z plane.
उत्तर
Given:
Electric field strength `vec"E" = 3/5 "E"_0 hat"i" + 4/5 "E"_0` \[ \stackrel\frown{j}\],
where E0 = 2.0 103 N/C
he plane of the rectangular surface is parallel to the y-z plane. The normal to the plane of the rectangular surface is along the x axis. Only `3/5 "E"_0`\[ \stackrel\frown{i}\], passes perpendicular to the plane; so, only this component of the field will contribute to flux.
On the other hand, `4/5 "E"_0`\[ \stackrel\frown{j}\] moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m2
Flux,
`phi = vec"E" . vec"a" = "E" xx "a"`
`phi = (3/5 xx 2 xx 10^3) xx ( 2 xx 10^-1) "N""m"^2 //"C"`
`phi = 0.24 xx 10^3 "N""m"^2//"C"`
`phi = 240 "N""m"^2//"C"`
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