Question

Define electric flux. Is it a scalar or a vector quantity?

A point charge \(q\) is kept at a distance of \(\dfrac{d}{2}\) directly above the centre of a square of side \(d\). Use Gauss' law to obtain the expression for the electric flux through this square.

Now if this point charge is moved to a point at a distance 'd' directly above the centre of the square and the side of the square is doubled, explain how the electric flux through the square will be affected.

Hint:

Imagine a cube of side d with the charge at its centre. Each face gets one sixth of \(q/\varepsilon_0\). Doubling the side and the height keeps the same shape.

Answer 1

Step 1: Define electric flux:
Electric flux through a surface is the total number of electric field lines passing normally through it. For a field \(\vec{E}\) and a small area \(d\vec{A}\), \(\Phi = \int \vec{E} \cdot d\vec{A}\).
The dot product of two vectors is a number, so electric flux is a SCALAR quantity. Its SI unit is \(\text{N m}^2 \text{C}^{-1}\).

Step 2: Build a closed surface:
The charge \(q\) is at height \(d/2\) above the centre of a square of side \(d\). Take five more squares of side \(d\) to form a cube of side \(d\).
The charge then sits exactly at the centre of this cube, and the given square is one face.

Step 3: Apply Gauss law:
The total flux through the closed cube is \(q/\varepsilon_0\).
By symmetry all six faces get equal flux. So the flux through the given square is
\[ \Phi = \frac{1}{6} \cdot \frac{q}{\varepsilon_0} = \frac{q}{6\varepsilon_0} \]

Step 4: Effect of the change:
Now the charge is at height \(d\) above the centre, and the square has side \(2d\). Its height is again half the side, \(d = 2d/2\).
The charge is again at the centre of a cube, this time of side \(2d\), and the square is one face. By the same argument the flux is \(q/6\varepsilon_0\).
So the flux does NOT change. The shape of the arrangement is only scaled, and Gauss law involves only the charge.

Final Answer:
Flux through the square \(= q/(6\varepsilon_0)\). After the change the flux stays the same. \[ \boxed{\Phi = \frac{q}{6\varepsilon_0}\ \text{(unchanged in the second case)}} \]