A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
Q/ϵ 0
Understanding Electric Flux and Gauss's Law for a Charge in a Cube
This question asks us to find the total electric flux passing through all six surfaces of a cube when a charge Q is placed exactly at its center. This is a classic problem that can be solved using Gauss's Law, a fundamental principle in electrostatics.
What is Electric Flux?
Electric flux is a measure of the total electric field passing through a given area. It tells us how many electric field lines penetrate a surface. Mathematically, for a small area element $d\vec{A}$, the flux is $\vec{E} \cdot d\vec{A}$, and the total flux through a surface is the integral of this quantity over the entire surface.
Applying Gauss's Law
Gauss's Law provides a powerful way to calculate electric flux, especially when dealing with symmetric charge distributions. Gauss's Law states that the total electric flux through any closed surface (called a Gaussian surface) is proportional to the total electric charge enclosed within that surface.
The mathematical expression for Gauss's Law is:
$$ \Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0} $$Where:
- $\Phi_E$ is the total electric flux through the closed surface.
- $\oint_S \vec{E} \cdot d\vec{A}$ is the surface integral of the electric field $\vec{E}$ over the closed surface S.
- $Q_{enclosed}$ is the total electric charge enclosed within the surface S.
- $\varepsilon_0$ is the permittivity of free space, a fundamental constant.
Solving the Problem: Charge at the Center of a Cube
In this problem, we have a charge Q placed at the center of a cube. The cube itself serves as our closed Gaussian surface. According to Gauss's Law, the total electric flux through this closed surface (the cube) depends only on the total charge enclosed within it.
- The closed surface is the cube with six faces.
- The charge enclosed within the cube is Q, as it's placed at the very center.
Using Gauss's Law, the total electric flux ($\Phi_{total}$) through the entire surface of the cube is given by:
$$ \Phi_{total} = \frac{Q_{enclosed}}{\varepsilon_0} $$Substituting the value of $Q_{enclosed}$ which is Q, we get:
$$ \Phi_{total} = \frac{Q}{\varepsilon_0} $$The question asks for the flux of the electric field through the six surfaces of the cube. The flux through the six surfaces combined is the total flux through the cube, which we just calculated using Gauss's Law.
Therefore, the total flux through the six surfaces of the cube is $\frac{Q}{\varepsilon_0}$.
Analysis of Options
Let's look at the given options:
- $Q/6\varepsilon_0$: This would be the flux through a single face if the charge were at the center, due to symmetry. However, the question asks for the flux through the six surfaces (total flux).
- $6Q/\varepsilon_0$: This value is 6 times the total flux, which is incorrect.
- $Q/\varepsilon_0$: This matches our calculation for the total flux through the entire cube surface using Gauss's Law.
- $5Q/6\varepsilon_0$: This value is arbitrary and does not follow from Gauss's Law for the total flux.
Based on Gauss's Law, the total electric flux through the closed surface of the cube, which encloses the charge Q, is $\frac{Q}{\varepsilon_0}$. This represents the flux through all six faces combined.
Revision Table: Key Concepts for Electric Flux and Gauss's Law
| Concept | Description | Formula/Principle |
|---|---|---|
| Electric Flux ($\Phi_E$) | Measure of electric field lines passing through a surface. | $\Phi_E = \int \vec{E} \cdot d\vec{A}$ |
| Gauss's Law | Relates total flux through a closed surface to the enclosed charge. | $\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$ |
| Gaussian Surface | An imaginary closed surface used to apply Gauss's Law. | Must be closed and enclose the charge. |
| Permittivity of Free Space ($\varepsilon_0$) | A constant representing the ability of a vacuum to permit electric fields. | Value is approximately $8.854 \times 10^{-12} \, C^2 N^{-1} m^{-2}$. |
Additional Information on Electric Flux and Gauss's Law Applications
- Symmetry is Key: Gauss's Law is most useful when the charge distribution has high symmetry (like spherical, cylindrical, or planar symmetry) because it allows us to easily evaluate the surface integral.
- Flux Through a Single Face: If the charge Q is at the center of a cube, due to the perfect symmetry, the electric flux is distributed equally among the six identical faces. So, the flux through a single face would indeed be $\frac{1}{6}$ of the total flux, which is $\frac{Q}{6\varepsilon_0}$. However, the question specifically asks for the flux through the "six surfaces," implying the total flux.
- Charge Position Matters (for individual faces): If the charge is *not* at the center (e.g., at a corner, edge, or outside), calculating the flux through individual faces becomes much more complex, often requiring direct integration or careful consideration of what fraction of the solid angle is subtended by each face relative to the charge position. However, the *total* flux through the closed cube surface still depends only on the total charge enclosed, regardless of its position inside (as long as it's inside). If the charge were outside the cube, the total flux through the cube would be zero.
- Applications: Gauss's Law is used to derive electric fields for various charge configurations, such as infinite lines of charge, infinite sheets of charge, and uniformly charged spheres.
Important Questions from Applications of Gauss’s Law
- The electric field lines from an isolated positively charged conducting sphere are
- An infinitely long straight uniformly charged wire has a linear charge density of $\lambda$. Calculate the work done by the electric field when a point charge $q$ is moved from an initial distance $r_1$ to a final distance $r_2$ ($r_2 > r_1$) from the wire.
The variation of electric field with respect to distance from centre of a charged conducting spherical shell of radius R is given by :
An infinitly long wire is charged uniformly with charge density λ and placed in air, the electric field at distance r from wire will be:
According to Gauss’s law, the electric field due to an infinitely long thin charged wire varies as:
