Consider a uniform electric field $\mathrm{E}==3\times {10}^3\stackrel{⏜}{i}\mathrm{N}/\mathrm{C}$. What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz-plane?

detailed solution

Correct option is A

Given that the electric field intensity is (3 x103) hence, the magnitude of electric field intensity is |E→|=3×103N/C, the side of the square is s=10cm=0.1m; area of the square is A=S2=0.01m2 The plane of the square is parallel to the y-z plane; therefore, the angle between the unit vector normal to the plane and electric field is given by θ = 0o. Now, the flux through the plane is given byϕ=|E→|Acos⁡θ……………….(1)Substituting the values in Eq. (1), we getϕ=3×103×0.01×cos⁡0∘=30Nm2/C.