Question

The direction of vector $A$ is radially outward from the origin, with $|A|=K{r}^n$ where $r^2=x^2+y^2+z^2$ and $K$ is constant. The value of $n$ for which $▽.A=0$ is

• A. $-2$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $-2$

If $f\to f\left(r\right),$ then using standard result
$div\overrightarrow{f}=\overrightarrow{▽}.\overrightarrow{f}=\frac{1}{r^2}\left(\frac{\partial }{\partial r}{r}^2|\overrightarrow{f}|\right)$
$\because |\overrightarrow{A}|=k{r}^n$ (given),
so $▽.\overrightarrow{A}=0$
$\Rightarrow \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2|A|\right)=0$
$\Rightarrow \frac{1}{r^2}\frac{\partial }{\partial r}(K.r^{n+2})=0$
$\Rightarrow \frac{K(n+2)}{r^2}.r^{n+1}=0$
$\Rightarrow K(n+2)r^{n-1}=0$
$n=-2$