Question

If $\varphi =\frac{1}{A},$ then $▽\varphi =$

• A. $\frac{A}{r}$
• B. $\frac{A}{r^2}$
• C. $\frac{A}{r^3}$
• D. $-\frac{A}{r^3}$

Answer 1

The correct option is D $-\frac{A}{r^3}$

$▽\varphi =▽\left(\frac{1}{r}\right)=▽\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)=▽\left\{{(x^2+y^2+z^2)}^{\frac{-1}{2}}\right\}$

$=i\frac{\partial }{\partial x}{(x^2+y^2+z^2)}^{-1/2}+j\frac{\partial }{\partial y}{(x^2+y^2+z^2)}^{-1/2}+k\frac{\partial }{\partial x}{(x^2+y^2+y^2)}^{-1/2}$

$=i\{-\frac{1}{2}{(x^2+y^2+z^2)}^{-3/2}2x\}+j\{-\frac{1}{2}{(x^2+y^2+z^2)}^{-3/2}2y\}+k\{-\frac{1}{2}{(x^2+y^2+z^2)}^{-3/2}2z\}$

$=\frac{-xi-yj-zk}{{(x^2+y^2+z^2)}^{-3/2}}=-\frac{A}{r^3}.$