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Question

According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:

The correct answer is

Q

Understanding Gauss's Law is fundamental in electromagnetism, particularly when dealing with electric fields and charge distributions. This law provides a powerful way to relate the electric flux through a closed surface to the charge enclosed within that surface.

Gauss's Law and Electric Flux Density

Gauss's Law can be stated in terms of both the electric field ($\mathbf{E}$) and the electric flux density ($\mathbf{D}$). The question specifically asks about the electric flux density, $\mathbf{D}$.

Electric flux density, $\mathbf{D}$, is related to the electric field, $\mathbf{E}$, by the equation:

\(\mathbf{D} = \varepsilon \mathbf{E}\)

where \(\varepsilon\) is the permittivity of the medium. In free space (vacuum), this is \(\varepsilon_0\).

Gauss's Law, in terms of electric flux density $\mathbf{D}$, states that the total electric flux of $\mathbf{D}$ passing through any closed surface is equal to the total electric charge enclosed within that surface.

Mathematically, this is expressed as the surface integral:

\(\oint_S \mathbf{D} \cdot d\mathbf{S} = Q_{enclosed}\)

Here:

  • \(\oint_S\) represents the integral over a closed surface S.
  • \(\mathbf{D}\) is the electric flux density vector.
  • \(d\mathbf{S}\) is an infinitesimal area vector element of the surface S, pointing outward.
  • \(\mathbf{D} \cdot d\mathbf{S}\) is the dot product, representing the normal component of $\mathbf{D}$ multiplied by the area element. The surface integral of this dot product gives the total outward flux of $\mathbf{D}$ through the surface.
  • \(Q_{enclosed}\) is the total charge enclosed within the closed surface S.

Analyzing the Question

The question asks for the value of the surface integral of the normal component of electric flux density D over a closed surface containing charge Q. According to Gauss's Law for $\mathbf{D}$, this surface integral is precisely equal to the total charge enclosed by the surface.

Given that the closed surface contains a total charge Q, the surface integral is:

\(\oint_S \mathbf{D} \cdot d\mathbf{S} = Q\)

Comparing this result with the given options:

  1. \(\frac{Q}{{{\varepsilon _0}}}\) - This would be the result for the flux of the electric field \(\mathbf{E}\) in free space, i.e., \(\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_{enclosed}}{{\varepsilon _0}}\). It is not the flux of \(\mathbf{D}\).
  2. \(\varepsilon _0 Q\) - This is not a form of Gauss's Law.
  3. Q - This directly matches the statement of Gauss's Law for electric flux density \(\mathbf{D}\).
  4. \(\frac{{{Q^2}}}{{{\varepsilon _0}}}\) - This is not a form of Gauss's Law.

Therefore, according to Gauss's Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is equal to Q.

Revision Table: Gauss's Law and Related Concepts

Concept Description Formula (Integral Form)
Electric Field (\(\mathbf{E}\)) Force per unit charge experienced by a test charge. Depends on source charges and the medium. \(\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_{enclosed}}{\varepsilon}\) (where \(\varepsilon\) is permittivity of the medium)
Electric Flux Density (\(\mathbf{D}\)) Related to \(\mathbf{E}\) by permittivity. Primarily depends only on the source charges, making it useful for dealing with different materials. \(\oint_S \mathbf{D} \cdot d\mathbf{S} = Q_{enclosed}\)
Permittivity (\(\varepsilon\)) A measure of how an electric field affects, and is affected by, a dielectric medium. \(\varepsilon = \varepsilon_r \varepsilon_0\). N/A
Permittivity of Free Space (\(\varepsilon_0\)) Permittivity of vacuum, a fundamental constant. \(\approx 8.854 \times 10^{-12}\) F/m
Gauss's Law Relates the electric flux through a closed surface to the enclosed charge. \(\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_{enclosed}}{\varepsilon}\) or \(\oint_S \mathbf{D} \cdot d\mathbf{S} = Q_{enclosed}\)

Additional Information on Electric Flux Density and Gauss's Law

The electric flux density $\mathbf{D}$ is often called the electric displacement field. While $\mathbf{E}$ depends on both the free charges (source charges) and any induced charges in dielectric materials within the medium, $\mathbf{D}$ is defined in a way that it only depends on the free charges.

This property makes the Gauss's Law involving $\mathbf{D}$ particularly useful when dealing with materials, such as dielectrics. The surface integral \(\oint_S \mathbf{D} \cdot d\mathbf{S}\) always equals the total free charge enclosed, regardless of the dielectric properties of the medium inside the surface.

In contrast, the surface integral of $\mathbf{E}$, \(\oint_S \mathbf{E} \cdot d\mathbf{S}\), equals the total charge enclosed (which includes both free and bound/induced charges) divided by the permittivity of the medium at each point on the surface. When the medium is uniform and linear, this simplifies to \(\frac{Q_{enclosed}}{\varepsilon}\), where \(Q_{enclosed}\) here means the total charge, not just free charge, or more precisely, the free charge only if the permittivity \(\varepsilon\) is uniform inside the surface. The $\mathbf{D}$ form is simpler and more universally applicable when considering free charges in various media.

Gauss's Law is one of Maxwell's equations, which are the fundamental equations governing classical electromagnetism.

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Important Questions from Electrostatics

  1. Electric potential $V$, is a ____ field, and electric field intensity $E$, is a ______ field.

  2. The dielectric constant of a vacuum is _____.

  3. What is the magnitude of the electric field at a distance $r$ from a point charge $Q$?
  4. The electric potential at the surface of an atomic nucleus (z = 50) of radius 9 × 10-15 m is ______________.

  5. What is the relative permittivity of slate dielectric?

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