Given a uniform electric field ` oversetto E =5xx10 ^(3) hati NC^(-1) ` . Find the flux of this field through 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?

Explanation

Here `oversetto E= 5xx 10^(3) hati NC^(-1) ` and area of square S = 10 `xx 10 cm ^(2) = 100 cm ^(2) = 100 xx 10^(4) m^(2) =0.01 m ^(2) `
Since surface is in Y-Z plane , hence `oversetto S =0.01 hat I m^(2) `
` therefore " " ` Flux ` phi_in oversetto E. oversetto S= ES =( 5 xx10^(3) ) xx0.01 = 50 N m^(2) C^(-1) `
When plane of square makes ` 30 ^(@) ` angle with the x-axis ,then area vector ` oversetto S ` subtends an angle ` theta = 60^(@) ` from x-axis and hence new flux will be
`" " phi_in . =ES cos theta = (5xx10^(3) ) xx 0.01 xx cos 60^(@) =25 N m^(2) C^(-1) `