Question

The condition that the vector \(\vec{A}\) should be a gradient of a scalar function is:

• A. \(\vec{\nabla} \cdot \vec{A} = 0\)
• B. \(\vec{\nabla}\, \vec{A} = 0\)
• C. \(\vec{\nabla} \times \vec{A} = 0\)
• D. \(\vec{\nabla} \times \vec{A} - \nabla^2 \vec{A} = 0\)

Hint:

The curl of any gradient is identically zero, so a gradient field must be irrotational.

Answer 1

The Correct Option is C

Step 1: Suppose \(\vec{A}\) is the gradient of a scalar function \(\phi\), so \(\vec{A} = \vec{\nabla}\phi\). We want the condition this places on \(\vec{A}\).

Step 2: Take the curl of both sides: \(\vec{\nabla} \times \vec{A} = \vec{\nabla} \times (\vec{\nabla}\phi)\).

Step 3: A standard vector identity states that the curl of any gradient is identically zero: \[\vec{\nabla} \times (\vec{\nabla}\phi) = \vec{0}.\] This holds because the second-order mixed partial derivatives of \(\phi\) commute.

Step 4: Therefore, if \(\vec{A}\) is a gradient it must be curl-free (irrotational). Conversely, a curl-free field on a simply connected region can always be written as a gradient, so this is the required condition: \[\boxed{\vec{\nabla} \times \vec{A} = 0}\]