Question

If \(\vec r=x\hat i+y\hat j+z\hat k\) and \(r=|\vec r|\), then the value of \(\operatorname{div}\left(\dfrac{\vec r}{r^3}\right)\) is

• A. \(1\)
• B. \(-1\)
• C. \(0\)
• D. \(2\)

Hint:

Remember the standard identity \(\nabla\cdot\left(\frac{\vec r}{r^3}\right)=0\) for \(r\neq 0\).

Answer 1

The Correct Option is C

Concept:
For the position vector \[ \vec r=x\hat i+y\hat j+z\hat k \] and \[ r=\sqrt{x^2+y^2+z^2} \] a standard vector identity is \[ \nabla\cdot\left(\frac{\vec r}{r^3}\right)=0 \] for \[ r\neq 0 \]

Step 1: Write the vector field. \[ \vec F=\frac{\vec r}{r^3} \] \[ \vec F=\frac{x\hat i+y\hat j+z\hat k}{(x^2+y^2+z^2)^{3/2}} \]

Step 2: Use the standard divergence result. The vector field \[ \frac{\vec r}{r^3} \] is an inverse square radial field. Its divergence is zero everywhere except at the origin. So, \[ \nabla\cdot\left(\frac{\vec r}{r^3}\right)=0 \]

Step 3: Final answer. \[ \boxed{0} \]