Question

A disk of radius \(a / 4\) having a uniformly distributed charge \(6 \mathrm{C}\) is placed in the \(x-y\) plane with its centre at \((-a / 2,0,0) .\) A rod of length a carrying a uniformly distributed charge \(8 \mathrm{C}\) is placed on the \(x\) -axis from \(x=a / 4\) to \(x=5 a / 4\). Two point charges \(-7 \mathrm{C}\) and \(3 \mathrm{C}\) are placed at \((a / 4,-a / 4,0)\) and \((-3 a / 4,3 a / 4,0)\), respectively. Consider a cubical surface formed by six surfaces \(x=\pm a / 2, y=\pm a / 2\), \(z=\pm a / 2 .\) The electric flux through this cubical surface is (A) \(\frac{-2 \mathrm{C}}{\varepsilon_{\mathrm{D}}}\) (B) \(\frac{2 \mathrm{C}}{c_{0}}\) (C) \(\frac{10 \mathrm{C}}{\varepsilon_{0}}\) (D) \(\frac{12 \mathrm{C}}{\varepsilon_{0}}\)

Short answer

\(\frac{3 \mathrm{C}}{\varepsilon_{0}}\)

Step-by-step solution

Understanding the Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. The law is stated as \(abla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}\), where \(\rho\) is the charge density, \(\vec{E}\) is the electric field, and \(\varepsilon_0\) is the permittivity of free space. The integral form of Gauss's Law for a closed surface (Gaussian surface) is \(\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\), where \(\Phi_E\) is the electric flux through the surface, \(d\vec{A}\) is the vector area of an infinitesimal piece of the surface, and \(Q_{\text{enc}}\) is the total charge enclosed by the surface.

Determining Which Charges are Enclosed

We need to determine which charges are enclosed by the cubical surface mentioned in the problem. Considering each object: the disk's centre, not falling into the cube, means its charge will not contribute to the flux through this cube. The rod's charge, which lies entirely outside the cube, does not contribute either. The point charge of \(-7 \mathrm{C}\) is outside the cube and hence does not contribute. The only charge enclosed by the cube is the point charge of \(3 \mathrm{C}\).

Calculating the Electric Flux

Since only the point charge of \(3 \mathrm{C}\) falls inside the cube, we can directly use Gauss's Law to determine the electric flux. According to Gauss's Law, \(\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}\). Substituting the value of the enclosed charge of \(3 \mathrm{C}\) into the equation, we get \(\Phi_E = \frac{3 \mathrm{C}}{\varepsilon_0}\).

Finding the Correct Option

Based on the calculated electric flux, the correct answer corresponds to the expression \(\frac{Q_{\text{enc}}}{\varepsilon_0}\) with the value of \(3 \mathrm{C}\). Compare this result with the given options to determine the correct choice.

Key concepts

Gauss's Law

Understanding Gauss's Law is essential when dealing with electric fields and charge distributions. It is a fundamental relationship that connects electric flux through a closed surface with the net charge within that surface.

Simply put, Gauss's Law states that the total electric flux through a closed surface, also known as a Gaussian surface, is directly proportional to the charge enclosed by the surface. The law can be expressed mathematically as \[\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\] where \(\Phi_E\) represents the electric flux, \(\vec{E}\) the electric field, \(d\vec{A}\) an infinitesimal area on the Gaussian surface with an outward normal direction, \(Q_{\text{enc}}\) the total charge enclosed, and \(\varepsilon_0\) the permittivity of free space. For a symmetric charge distribution, this law can greatly simplify the process of finding the electric field.

Electric Field

The electric field, denoted as \(\vec{E}\), is a vector field around charged particles or objects. It represents the force per unit charge a positive test charge would experience in the vicinity of the source charge.

The magnitude and direction of the electric field are influenced by the charge distribution. It decreases with distance from the source charge in the case of a point charge and can be more complex for distributed charges. The electric field can be calculated from Gauss's Law for symmetrical distributions like spherical, cylindrical, and planar.

Charge Density

Charge density, usually symbolized as \(\rho\), is a measure of the amount of electric charge per unit volume of space, in the case of volume charge density, or per unit area or length in the case of surface or line charge density, respectively.

When solving problems using Gauss's Law, knowing the type of charge distribution is necessary — whether it's a volume, surface, or line distribution. Each type affects the electric field differently. In our textbook problem, we dealt with different configurations, such as a uniformly charged disk and a rod, which correspond to surface and line charge densities, respectively.

Permittivity of Free Space

The permittivity of free space, also known as the electric constant, is denoted by the symbol \(\varepsilon_0\). It is a fundamental physical constant which is the measure of the ability of the vacuum to permit the electric field lines. This constant appears in Coulomb's law as well as Gauss's Law and is critical in determining the strength of the electric field generated by a given charge distribution.

The value of \(\varepsilon_0\) in SI units is approximately \(8.85 \times 10^{-12}\) farads per meter (F/m). In our exercise, it helps us determine the electric flux through the Gaussian surface by relating it to the enclosed charge.