Question

For electrostatic fields in charge free atmosphere, which one of the following is correct?

• A. \( \nabla \times E = 0 \text{ and } \nabla \cdot E = 0 \)
• B. \( \nabla \times E \neq 0 \text{ and } \nabla \cdot E = 0 \)
• C. \( \nabla \times E = 0 \text{ and } \nabla \cdot E \neq 0 \)
• D. \( \nabla \times E \neq 0 \text{ and } \nabla \cdot E \neq 0 \)

Hint:

An electrostatic field can never have any curl component (\(\nabla \times E = 0\)). This property distinguishes static configurations explicitly from time-varying fields where Faraday's induction introduces a non-zero curl.

Answer 1

The Correct Option is A

Understanding the Concept: Maxwell's equations govern the behavior of electromagnetic fields. For static electric fields (electrostatics):
• Conservative Property: An electrostatic field is irrotational and conservative, meaning the line integral of the field around any closed loop is identically zero: \[ \nabla \times E = 0 \]
• Gauss's Law: The divergence of the electric field is proportional to the local charge density \(\rho_v\): \[ \nabla \cdot E = \frac{\rho_v}{\varepsilon} \]

Step 1: Analyze under Charge-Free condition
In a charge-free atmosphere, the volume charge density is identically zero everywhere: \[ \rho_v = 0 \] Substituting \(\rho_v = 0\) into Gauss's differential law: \[ \nabla \cdot E = 0 \] Since the field is static and conservative regardless of the presence of charge: \[ \nabla \times E = 0 \] Combining both results yields the conditions \( \nabla \times E = 0 \) and \( \nabla \cdot E = 0 \), which matches Option (A).