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The electric field in a region is given ...

The electric field in a region is given by `vecE=3/5 E_0veci+4/5E_0vecj` with `E_0=2.0 xx 10^3 N C^(-1)`. Find the flux of this field through a recatngular surface of area `0.2m^2` parallel to the y-z plane.

Text Solution

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`vecE=3/5 E_0hati+4/5E_0hatj`
Where `E_0=2xx10^3` N/C
Area = `0.2m^2` parallel to YZ plane.
Normal to the area will be along x- axis electric flux is
`phi = E_xxxA`
`=3/5E_0xx0.2`
`=0.6/5xx2xx10^3`
Electric flux `phi = 240Nm^2C^-1`
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