The total charge enclosed in an incremental volume of 2 x 10^-9 m^3 located at the origin is ......

Q: Why do we use divergence here instead of the standard integral form of Gauss's Law?

The integral form of Gauss's Law ($\oint \vec{D} \cdot d\vec{A} = Q_{enc}$) is great when you have a large, symmetric closed surface. However, here we are given the flux density as a continuous vector field and asked for the charge in an *incremental* (infinitesimally small) volume at a specific point. The differential form ($\nabla \cdot \vec{D} = \rho_v$) is specifically designed to relate the field at a single point to the charge density at that exact point.

Q: What exactly does the 'incremental volume' $dV$ mean in this context?

An incremental volume $dV$ is a volume so small that any macroscopic property within it, like the charge density $\rho_v$, can be considered perfectly constant. This allows us to bypass complex volume integrals and simply use the algebraic multiplication $Q = \rho_v \cdot dV$.

Q: Why did the $x$ and $y$ terms completely cancel out when calculating the divergence?

That cancellation is a specific property of the vector field $\vec{D}$ provided in the problem. Physically, it means that whatever flux enters a tiny volume element along the x and y directions exactly equals the flux leaving it in those directions. The only net outward flow of flux (which indicates the presence of charge) is happening along the z-axis.

Q: What if the incremental volume was located at $(1, 2, 3)$ instead of the origin?

In this specific problem, the divergence evaluated to a constant value of $2 \text{ C/m}^3$ everywhere, independent of $x$, $y$, or $z$. Therefore, the charge density is uniform throughout all of space. The enclosed charge would still be $4 \text{ nC}$ regardless of where you place that tiny volume element.

Q: What is the difference between the electric field $\vec{E}$ and the electric flux density $\vec{D}$?

The electric flux density $\vec{D}$ is defined as $\vec{D} = \epsilon_0 \vec{E} + \vec{P}$, where $\vec{P}$ is the polarization of the medium. In free space (vacuum), this simplifies to $\vec{D} = \epsilon_0 \vec{E}$. The vector $\vec{D}$ is incredibly useful in electromagnetics because its divergence depends *only* on the free charge density, completely ignoring any complicated bound charges in dielectric materials.

Question

The total charge enclosed in an incremental volume of $2 	imes 10^{-9} 	ext{ m}^3$ located at the origin is ...... nC, if electric flux density of its field is found as $D = e^{-x} \sin y \hat{i} - e^{-x} \cos y \hat{j} + 2z \hat{k} 	ext{ C/m}^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a tiny, almost invisible box located exactly at the origin of our 3D coordinate system. This is our incremental volume element, $dV = 2 	imes 10^{-9} 	ext{ m}^3$.

We are also given the electric flux density vector, $\vec{D}$, which describes how the electric field lines are flowing through space. The equation for this vector field is:

$$ \vec{D} = e^{-x} \sin y \hat{i} - e^{-x} \cos y \hat{j} + 2z \hat{k} 	ext{ C/m}^2 $$

Our mission is to find the total charge $Q$ hidden inside this tiny box. To do this, we need a mathematical bridge that connects the flow of the field (flux density) to the source of the field (charge).

### The Master Equation: Gauss's Law in Differential Form

The perfect tool for this job is the **differential form of Gauss's Law**. While the integral form deals with large surfaces, the differential form zooms in on a single point. It states that the divergence of the electric flux density vector $\vec{D}$ is exactly equal to the volume charge density $ho_v$ at that specific point:

$$ 
abla \cdot \vec{D} = ho_v $$

The divergence operator $
abla \cdot$ tells us how much the field is "spreading out" from a point. If the divergence is positive, there is a source of flux (positive charge). If it's negative, there is a sink (negative charge).

### Executing the Divergence

Let's calculate the divergence by taking the sum of the partial derivatives of each component of $\vec{D}$ with respect to its corresponding spatial coordinate:

$$ ho_v = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z} $$

Substituting our specific components, we get:

$$ ho_v = \frac{\partial}{\partial x}(e^{-x} \sin y) + \frac{\partial}{\partial y}(-e^{-x} \cos y) + \frac{\partial}{\partial z}(2z) $$

Now, we carefully execute each derivative. Remember to treat the other variables as constants during partial differentiation:

$$ ho_v = -e^{-x} \sin y - e^{-x}(-\sin y) + 2 $$

Notice the beautiful mathematical symmetry here! The first two terms perfectly cancel each other out:

$$ ho_v = -e^{-x} \sin y + e^{-x} \sin y + 2 = 2 	ext{ C/m}^3 $$

We are left with a constant volume charge density of $2 	ext{ C/m}^3$. This means the charge density is uniform everywhere in space, including at our origin $(0,0,0)$.

### Final Calculation

Since we are dealing with an infinitesimally small incremental volume $dV$, we can assume the charge density $ho_v$ is constant throughout this tiny space. Therefore, the total enclosed charge $Q$ is simply the product of the charge density and the volume:

$$ Q = ho_v \cdot dV $$

Substituting our known values:

$$ Q = 2 	imes (2 	imes 10^{-9}) = 4 	imes 10^{-9} 	ext{ C} $$

Converting this to nano-coulombs (where $1 	ext{ nC} = 10^{-9} 	ext{ C}$), we arrive at our final, elegant answer:

$$ Q = 4 	ext{ nC} $$

Animated Solution for Physics - Electrostatics: The total charge enclosed in an incremental volume of $2 \times 1 0^{- 9} m^{3}$ located at the origin is ...... nC, if electric flux density of its field is found as $D = e^{- x} sin y \hat{i} - e^{- x} cos y \hat{j} + 2 z \hat{k} C/m^{2}$ .

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Given Data}

\text{Gauss's Law in Differential Form}

\text{Setting up the Divergence}

\text{Calculating the Divergence}

\text{Evaluating the Derivatives}

\text{Simplifying the Expression}

\text{Charge Density at the Origin}

\text{Calculating Total Enclosed Charge}

\text{Final Substitution and Result}

\text{What if the density wasn't constant?}

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing the Setup

The Master Equation

Executing the Divergence

Final Calculation

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