Given a uniform electric field `vecE=5xx10^3hati`N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis ?

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#### Solution

When the plane is parallel to the *y-z *plane:

Electric flux, `phi=vecE.vecA`

Here:

`vecE=5xx10^3hati "N/C"`

`vecA=(10cm^2)hati=10^(-2)hatim^2`

`:.phi=(5xx10^3hati).(10^(-2)hati)`

⇒ ϕ =50 Weber

When the plane makes a 30° angle with the* x*-axis, the area vector makes 60° with the *x*-axis.

`phi=vecE.vecA`

⇒ϕ=EA cosθ

⇒ϕ=(5×10^{3})(10^{−2})cos60°

`=>phi=50/2`

⇒ ϕ =25 Weber

Concept: Electric Flux

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