A charge q is placed at the centre of the base of a square pyramid. The net outward electric flux across each of the slanted faces is (Consider permittivity as \(\epsilon_0\))
Show Hint
For Gauss's law problems with open surfaces, the key is always to imagine a larger, closed symmetrical surface that includes the given open surface as a part. Place the charge at the center of this imaginary surface, calculate the total flux, and then divide it up based on symmetry to find the flux through the desired part.
Step 1: Understanding the Concept:
This problem requires finding the electric flux through a part of a surface that encloses a charge. The key tool for solving such problems is Gauss's Law, which relates the total electric flux through a closed surface to the net charge enclosed by it. Since the pyramid is an open surface, we must use a symmetry argument to construct a closed Gaussian surface. Step 2: Key Formula or Approach:
Gauss's Law states that the total electric flux \(\Phi_E\) through any closed surface is equal to the net charge enclosed \(Q_{enc}\) divided by the permittivity of free space \(\epsilon_0\).
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \]
We will use a symmetry argument by constructing a closed surface where the charge \(q\) is at a point of high symmetry. Step 3: Detailed Explanation:
1. Constructing a Closed Surface: The given surface is a square pyramid, which is an open surface. The charge \(q\) is located at the center of its square base. To apply Gauss's Law, we need a closed surface. We can create a symmetrical closed surface by placing an identical, inverted pyramid below the given pyramid, such that they share the same square base.
This construction forms a closed square bipyramid (which looks like an octahedron if the faces are equilateral triangles, though that's not required).
2. Applying Gauss's Law: The charge \(q\) is now at the geometric center of this closed bipyramid. According to Gauss's Law, the total electric flux \(\Phi_{\text{total}}\) emanating from the charge \(q\) through the entire closed surface of the bipyramid is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]
3. Using Symmetry: The bipyramid is composed of two identical pyramids (the original one and the inverted one). The charge \(q\) is symmetrically placed with respect to both pyramids. Therefore, the total flux must be shared equally between the upper pyramid and the lower pyramid.
The flux through all the slanted faces of the original (upper) pyramid, \(\Phi_{\text{pyramid}}\), is half of the total flux.
\[ \Phi_{\text{pyramid}} = \frac{1}{2} \Phi_{\text{total}} = \frac{1}{2} \frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0} \]
Note that there is no flux through the base of the pyramid because the charge lies in the plane of the base, so the electric field lines are parallel to the base surface, making \(\mathbf{E} \cdot d\mathbf{A} = 0\) for the base area.
4. Finding Flux through a Single Face: The question asks for the flux across \textit{each} of the slanted faces. The original pyramid has 4 identical slanted faces. Since the charge is at the center of the square base, the flux \(\Phi_{\text{pyramid}}\) is distributed equally among these 4 faces due to symmetry.
Therefore, the flux through a single slanted face, \(\Phi_{\text{face}}\), is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{pyramid}}}{4} = \frac{q/(2\epsilon_0)}{4} = \frac{q}{8\epsilon_0} \]
Step 4: Final Answer:
The net outward electric flux across each of the slanted faces is \(\frac{q}{8\epsilon_0}\). Therefore, option (D) is the correct answer.
