© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recognize the Relation Between D and Charge Density
The electric flux density vector \overline{D} is related to the free charge density \rho by the differential form of Gauss’s law:
\nabla \cdot \overline{D} = \rho \,.
Here, \overline{D} is given as:
\overline{D} = e^{-x} \sin y \,\hat{i} \;-\; e^{-x} \cos y \,\hat{j} \;+\; 2z\,\hat{k} \quad \text{(C/m}^2\text{)} \,.
Step 2: Compute the Divergence of the Given \overline{D}
The divergence \nabla \cdot \overline{D} is computed by taking partial derivatives of each component with respect to its corresponding variable:
\nabla \cdot \overline{D} = \frac{\partial}{\partial x}\bigl(e^{-x}\sin y\bigr) \;+\; \frac{\partial}{\partial y}\bigl(-\,e^{-x}\cos y\bigr) \;+\; \frac{\partial}{\partial z}\bigl(2z\bigr).
Let us evaluate each partial derivative:
\frac{\partial}{\partial x}\bigl(e^{-x}\sin y\bigr) = -\,e^{-x} \sin y.
\frac{\partial}{\partial y}\bigl(-\,e^{-x}\cos y\bigr) = -\,e^{-x}(-\,\sin y) = e^{-x}\sin y.
\frac{\partial}{\partial z}\bigl(2z\bigr) = 2.
Combining these:
\nabla \cdot \overline{D} = \bigl(-\,e^{-x}\sin y\bigr) \;+\; \bigl(e^{-x}\sin y\bigr) \;+\; 2 = 2.
Thus,
\nabla \cdot \overline{D} = 2 = \rho \,.
Step 3: Determine the Total Charge Enclosed
Since \rho (the free charge density) is constant and equal to 2 C/m3 at the origin, the total charge enclosed in a volume V is:
q = \rho \times V.
Given:
V = 2 \times 10^{-9}\,\text{m}^3 \quad \text{and} \quad \rho = 2\,\text{C/m}^3,
we have:
q = 2 \,\times\, 2 \times 10^{-9} = 4 \times 10^{-9}\,\text{C}.
Step 4: Convert the Charge to NanoCoulombs (nC)
Recall that 1\,\text{C} = 10^{9}\,\text{nC} . Therefore,
q = 4 \times 10^{-9}\,\text{C} = 4\,\text{nC}.
Final Answer
The total charge enclosed is 4 nC.
