Question

Let the electric field in some region $R$ be given by $\vec{E} = e^{-y^2}\hat{i} + e^{-x^2}\hat{j}$. From this we conclude that
 

• A. $R$ has a non-uniform charge distribution.
• B. $R$ has no charge distribution.
• C. $R$ has a time-dependent magnetic field.
• D. The energy flux in $R$ is zero everywhere.

Hint:

Whenever $\nabla\cdot\vec{E}\neq 0$, charge must be present.

Answer 1

The Correct Option is B, C

Step 1: Use Gauss's law in differential form.
\[ \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} \] Compute divergence: \[ \nabla \cdot \vec{E} = \frac{\partial}{\partial x}(e^{-y^2}) + \frac{\partial}{\partial y}(e^{-x^2}) \] Both terms are nonzero because derivatives of exponentials exist. Thus $\nabla \cdot \vec{E} \neq 0$.

Step 2: Interpretation.
If divergence is nonzero, charge density is nonzero. Thus region contains a non-uniform charge distribution.

Step 3: Conclusion.
Correct: (A).