Problem 63
Question
Suppose, an imaginary cube is with a charge situated at the centre of it. The total electric flux passing through each of the faces of the cube will be (a) \(\frac{q}{6 \varepsilon_{0}}\) (b) \(\frac{q}{2 \varepsilon_{0}}\) (c) \(\frac{q}{12 \varepsilon_{0}}\) (c) None of these
Step-by-Step Solution
Verified Answer
The flux through each face of the cube is \(\frac{q}{6 \varepsilon_{0}}\).
1Step 1: Understanding the Problem
We are given a scenario with a charge located at the center of an imaginary cube. The problem asks for the total electric flux passing through each face of the cube.
2Step 2: Understanding Electric Flux
The electric flux, represented as \(\Phi\), through a closed surface due to a point charge is given by Gauss's Law: \(\Phi = \frac{q}{\varepsilon_0}\), where \(q\) is the charge and \(\varepsilon_0\) is the vacuum permittivity.
3Step 3: Applying Gauss's Law to the Cube
Since the charge is at the center of the cube, the electric flux is evenly distributed through all the cube's faces. The total flux through the entire surface of the cube is \(\frac{q}{\varepsilon_0}\).
4Step 4: Finding the Flux Through One Face
A cube has 6 faces. Since the flux is uniformly distributed, the flux through each face is given by dividing the total flux by 6: \(\Phi_{face} = \frac{q}{6\varepsilon_0}\).
Key Concepts
Gauss's LawCubic SymmetryPoint Charge Distribution
Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism, connecting the electric flux emanating from a surface to the charge enclosed within that surface. It is mathematically expressed as \( \Phi = \frac{q}{\varepsilon_0} \), where \( \Phi \) represents the electric flux, \( q \) is the charge, and \( \varepsilon_0 \) is the vacuum permittivity. This law simplifies the calculation of electric fields when dealing with symmetrical charge distributions.When applying Gauss's Law, it is important to choose an appropriate closed surface, known as a Gaussian surface. The symmetry of the problem often guides this choice. For instance, when dealing with a point charge placed at the center of a cube, as in the exercise, the cube's surface serves as the Gaussian surface. The advantage of using Gauss's Law lies in its ability to bypass complex integration by exploiting symmetrical properties of electric fields.In simple terms, Gauss's Law allows us to calculate how much electric field (
- passes through an imaginary surface
- encloses a given charge distribution
Cubic Symmetry
Cubic symmetry refers to the situation where an object, such as a cube, and the charge distribution, such as a point charge at the center, have symmetrical properties. This symmetry can greatly influence the electric flux distribution across the surface of the cube.
In the context of a cube with a point charge at its center, the symmetry of the cube plays a crucial role. Due to this symmetry:
- The electric field lines emanating from the charge are spread evenly across all directions.
- This results in an equal distribution of electric flux through each face of the cube.
Point Charge Distribution
A point charge distribution refers to a single point where all the charge is concentrated. In our problem scenario, a point charge is located precisely at the center of an imaginary cube.
When dealing with point charges, one should:
- acknowledge that the electric field radiates outward symmetrically from the charge.
- realize that this is an idealized model, making the calculations easier under the assumption of a vacuum or infinite medium.
Other exercises in this chapter
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