Consider a uniform electric field `vecE=3xx10^(3)hatiNC^(-1)`
a. Whate is the flux of this field through a square of 0.1 m on a side whose plane is parallel to the y-z plane?
b. What is the flux through the same square if the normal to its plane makes a 60 angle with x-axis?

Explanation

Given `vecE=3xx10^(3)hatiNC^(-1),a=0.1m`
a. A normal drawn from P will coincide with `vecE`
using the formule for electric flux,
`phi=ES cos theta , theta =0^(@)`
i.e. `phi =#S=3xx10^(3)xx(0.1)^(2)Vm`
i.e. `phi=30Vm` (or `Nm^(2)C^(-1)`)
b. For `theta=60^(@)`
`phi=3xx10^(3)xx(0.1)^(2)xxcos60^(@)`
i.e `phi=15vm` (or `Nm^(2)C^(-1)`)