Question

The surface integral of the normal component of electric flux density over any closed surface is equal to the following enclosed:

• A. current
• B. charge
• C. voltage
• D. capacitance

Hint:

By keeping track of the units, you can easily verify this relationship: the unit of electric flux density $D$ is $\text{C/m}^2$. Integrating this over a surface area ($\text{m}^2$) gives a net unit of Coulombs ($\text{C}$), which is the standard SI unit for electrical charge.

Answer 1

The Correct Option is B

Concept: This question directly restates Gauss's Law for Electric Fields, which is the first of Maxwell's four equations. In integral form, Gauss's Law states that the net outward electric flux passing through any closed boundary surface is equal to the total net charge enclosed inside that volume. Mathematically, it is written as: \[ \oint_{S} \vec{D} \cdot d\vec{S} = Q_{\text{enclosed}} \] Where: - $\vec{D}$ is the electric flux density vector (expressed in Coulombs per square meter, $\text{C/m}^2$). - $d\vec{S}$ is an infinitesimal area element vector pointing normal to the surface. - $Q_{\text{enclosed}}$ is the net total electric charge inside the surface boundary.

Step 1: Analyze the question's phrasing.
The question statement text reads: "The surface integral of the normal component of electric flux density over any closed surface..." This corresponds exactly to the left-hand expression of Gauss's Law: \[ \oint_{S} \vec{D} \cdot \hat{n} \, dS \]

Step 2: Match with the fundamental physical quantity.
According to Gauss's law, this integral evaluates exactly to the total enclosed charge. This matches Option (B).