Consider a uniform electric field E = 3 × 10 3 î N/C. (a) What is the flux of this field through a square of 10 cm 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 the x-axis?
(a) Electric field intensity, E=3×103N/C Magnitude of electric field intensity, ∣E∣=3×103N/C Side of the square, s=10cm=0.1m Area of the square, A=s2=0.01m2 The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ=0o Flux (ϕ) through the plane is given by the relation, ϕ=∣E∣Acosθ =3×103×0.01×cos0o =30Nm2/C (b) Plane makes an angle of 60o with the x-axis. Hence, θ=60o Flux, ϕ=∣E∣Acosθ =3×103×0.01×cos60o =30×21=15Nm2/C
