Question

Find \( \nabla\phi \) if \( \phi = \log r \), where \( r = |\vec{r}| \):

• A. \( \dfrac{\vec{r}}{r} \)
• B. \( \dfrac{\vec{r}}{r^{2}} \)
• C. \( \dfrac{\vec{r}}{r^{3}} \)
• D. \( 0 \)

Hint:

Use \( \nabla f(r)=f'(r)\,\vec r/r \). With \( f=\log r \), \( f'(r)=1/r \), so the answer is \( \vec r/r^2 \).

Answer 1

The Correct Option is B

Step 1: Compute the gradient component by component from $\phi=\log r=\tfrac12\log(x^{2}+y^{2}+z^{2})$.

Step 2: The $x$-component:
\[ \frac{\partial \phi}{\partial x} = \frac{1}{2}\cdot\frac{2x}{x^{2}+y^{2}+z^{2}} = \frac{x}{r^{2}}. \]

Step 3: By identical reasoning the $y$- and $z$-components are $\dfrac{y}{r^{2}}$ and $\dfrac{z}{r^{2}}$.

Step 4: Assemble the vector:
\[ \nabla\phi = \frac{x\hat{i}+y\hat{j}+z\hat{k}}{r^{2}} = \frac{\vec{r}}{r^{2}}. \]

Step 5: This is the radial field falling off as $1/r$ in magnitude (since $|\vec r|=r$), confirming $\nabla(\log r)=\vec{r}/r^{2}$.
\[ \boxed{\, \nabla\phi = \dfrac{\vec{r}}{r^{2}} \,} \]