The direction of vector A is radially outward from the origin, with |A| = krn where r2 = x2 + y2 + z2 and k is a constant. The value of n for which $$\nabla \cdot {\text{A}} = 0$$ is A. -2 B. 2 C. 1 D. 0

-2
2
1
0

The correct answer is $\boxed{0}$.

The divergence of a vector field is defined as the sum of the partial derivatives of the field’s components with respect to the coordinates. In this case, the vector field is $\text{A} = krn$, where $r$ is the radial distance from the origin and $k$ is a constant. The partial derivatives of $r^n$ with respect to $x$, $y$, and $z$ are $nr^{n-1}x$, $nr^{n-1}y$, and $nr^{n-1}z$, respectively. Therefore, the divergence of $\text{A}$ is

$$\nabla \cdot \text{A} = \frac{\partial}{\partial x} (kr^n) + \frac{\partial}{\partial y} (kr^n) + \frac{\partial}{\partial z} (kr^n) = nkr^{n-1}.$$

Setting this equal to zero and solving for $n$, we find that $n=0$.

The other options are incorrect because they do not satisfy the equation $\nabla \cdot \text{A} = 0$. For example, if $n=-2$, then the divergence of $\text{A}$ is $2kr^{-1}$, which is not equal to zero.

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