If electric field in the space is given by, `vec E = 4x hati + (y^2+1)hatj` and electric flux through ABCD is `phi_1` and electric flux through BCEF is `phi_2`, then find (`phi_1 - phi_2`)

Mathematically, the electric flux (phi) through small area ds placed in electric field E is given by

Electric potential V and electric flux phi are

The electric field in a region of space is given by E=5hati+2hatjN//C . The flux of E due ot this field through an area 1m^2 lying in the y-z plane, in SI units is

The electric field in a region of space is given by E=5hati+2hatjN//C . The flux of E due ot this field through an area 1m^2 lying in the y-z plane, in SI units is

In a region, the intensity of an electric field is given by vecE=2hati+3hatj+hatk in NC^(-1) . The electric flux through a surface vecS=10hatim^(2) in the region is

An electric field vecE=4xhati-(y^(2)+1)hatjN//C passes through the box shown in figure. The flux of the electric field through surface ABCD and BCGF are marked as phi_1)and phi_(11) respecticvely. The differentce between (phi_(1)-phi_(1)) is in Nm^(2) /C)

An electric field vecE=4xhati-(y^(2)+1)hatjN//C passes through the box shown in figure. The flux of the electric field through surface ABCD and BCGF are marked as phi_1and phi_(2) respectively. The difference between (phi_(1)-phi_(2)) is in Nm^(2) /C)

Electric flux phi is a vector quantity.