The law goverening electrostatics is coulomb's law.In priciple coulomb's law can be used to compute electric field due to any charge configuation .In pracitce however it is a formidable task to compute electric field due to uniform charge distributions .For such cases Gauss proposed a theorem which states that
`ointbar(E).bar(ds)=(q)/(epsilon_(0))`
where `ds` is an element of area directed across the outward normal for the surface at every point .The integral is called electric flux.Any convenient surface that we choose to evalute the surface integral is called the Gaussian surface.
Consider a uniform electric field `E=3xx10^(3)hat(i)NC^(-1)` what is the flux thorugh a square of side `10` cm whose plane is parallel to the `y-z` plane?

Options

A. `30NC^(-1)m^(2)`
B. `20NC^(-1)m^(2)`
C. `10NC^(-1)m^(2)`
D. zero

Answer

A

Explanation

`phi=E Scostheta` theta is angle between area vector and field vector `S=10cmxx10cm=10^(-2)m^(2)`
`E=3xx10^(3)NC^(-1):. Phi=Escostheta=30NC^(-1)m^(2)`