A point charge + 10 μ C is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? ( Hint: Think of the square as one face of a cube with edge 10 cm.)

Question

Philoid · Accepted Answer

Given:

q = + 10

s = 10 cm

Assume the charge to be enclosed by a cube, where the square is one of its sides.

Now, let us find the total flux through the imaginary cube.

We know that,

Flux, Φ = q/ε …(1)

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Now plugging the values of q and ε₀ in equation (2)

⇒ Φ = 11.28 × 10⁵ Nm^-2C^-1

We understand that flux through all the faces of cube will be equal;

Let flux through the square = Φ_a

Hence,

Φ_a = Φ/6

Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.

Φ_a = 1.88 Nm^-2C^-1

The flux through the square is.

A point charge + 10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)