Question

A point charge \(q\) is placed at a distance \(a/2\) directly above the center of a square of side \(a\). The electric flux through the square is:

• A. \(\frac{q}{\varepsilon_0}\)
• B. \(\frac{q}{6\varepsilon_0}\)
• C. \(\frac{q}{4\varepsilon_0}\)
• D. \(\frac{q}{2\varepsilon_0}\)

Hint:

Whenever a charge is placed at a perpendicular distance \(a/2\) from the center of a square of side \(a\), always construct a cube around it to use symmetry. The flux through the single face will be \(1/6\) of the total flux.

Answer 1

The Correct Option is B

Step 1: Read the situation carefully.
A point charge $q$ sits at a height $a/2$ straight above the centre of a flat square of side $a$. We want the electric flux that pierces through this one square. Flux through an open surface is awkward to compute directly, so we will use symmetry instead.
Step 2: Recall Gauss's Law.
Gauss's Law tells us the total flux leaving any closed surface depends only on the charge trapped inside it: \[ \Phi_{\text{total}} = \frac{q_{\text{enc}}}{\varepsilon_0} \] The trick is to wrap our single square into a neat closed box.
Step 3: Build a cube around the charge.
Notice the height $a/2$ is exactly half the side $a$. So if we imagine a cube of side $a$ and place our square as the bottom face, the charge at height $a/2$ lands precisely at the dead centre of that cube.
Step 4: Apply Gauss's Law to the full cube.
The charge $q$ is now fully enclosed by this cube, so the total flux through all six faces together is \[ \Phi_{\text{total}} = \frac{q}{\varepsilon_0} \]
Step 5: Use symmetry to share the flux.
Sitting at the exact centre, the charge is equally distant from each of the six identical faces. By symmetry the flux splits evenly, so each face gets one sixth of the total.
Step 6: Pick out our one square.
Our square is just one of those six faces, hence \[ \Phi_{\text{square}} = \frac{1}{6}\,\frac{q}{\varepsilon_0} = \frac{q}{6\varepsilon_0} \] This matches option (B).
\[ \boxed{\Phi = \dfrac{q}{6\varepsilon_0}} \]