Question

Consider a uniform electric field \(E = 3 × 10^3 \,\hat{i} NC^{-1}\). 
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz - plane? 
(b) What is the flux through the same square if the normal to its plane makes a \( 60^0\) angle with the x-axis?

Answer 1

(a) Electric field intensity,\(E = 3 × 10^3 \,\hat{i} NC^{-1}\)
Magnitude of electric field intensity, \(|E| = 3 × 10^3 NC^{-1}\)
Side of the square, \(s = 10 cm = 0.1 m\)
Area of the square, \(A = s^2 = 0.01 m^2\)
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, \(\theta = 0^0\)
Flux (Φ ) through the plane is given by the relation,
\(Φ = |E| Acos θ\)
\(= 3 × 10^3 × 0.01 × cos 0^0\)
\(= 30 Nm^2C^{-1}\)

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(b) Plane makes an angle of \(60^0\) with the x – axis. Hence, \(θ = 60^0\)
\(Flux, Φ = |E| Acos θ\)
\(= 3 × 10^3 × 0.01 × cos 60^0\)
\(= 30 ×\frac{1}{2}\)
\(= 15 N m^2C^{-1}\)