Question

Q- Divergence of r / r3 is (a) zero at the origin (b) zero everywhere (c) zero everywhere except the origin (d) nonzero everywhere

Answer 1

Answer:

the answer is C I think

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO FIND THE CORRECT OPTION

$\displaystyle \sf{ \nabla \bigg( \frac{ \vec{r}}{ {r}^{3} } \bigg) } \: \: \: \: is$

(a) zero at the origin

(b) zero everywhere

(c) zero everywhere except the origin

(d) nonzero everywhere

CALCULATION

Here

$\sf{ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \: }$

So

$\displaystyle \: \sf{ r = | \vec{r}| = \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} } = \: {({x}^{2} + {y}^{2} + {z}^{2})}^{ \frac{1}{2} } \: }$

$\implies \displaystyle \sf{ {r}^{3} = \: {({x}^{2} + {y}^{2} + {z}^{2})}^{ \frac{3}{2} } \: }$

So

$\displaystyle \sf{ \frac{ \vec{r}}{ {r}^{3} } \: = \: {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \times (x \hat{i} + y \hat{j} + z \hat{k} ) }$

So

$\displaystyle \sf{ \nabla \bigg( \frac{ \vec{r}}{ {r}^{3} } \bigg) }$

$= \displaystyle \sf{ \frac{ \partial}{ \partial x} \bigg [ x{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] + \frac{ \partial}{ \partial y} \bigg [ y{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] \: +\frac{ \partial}{ \partial z} \bigg [ z{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] \: }$

Now

$\displaystyle \sf{ \frac{ \partial}{ \partial x} \bigg [ x{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] }$

$= \displaystyle \sf{ \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] + x.2x. \frac{ - 3}{2} \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

$= \displaystyle \sf{ \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 {x}^{2} \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

So

$\displaystyle \sf{ \frac{ \partial}{ \partial x} \bigg [ x{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] } = \displaystyle \sf{ \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 {x}^{2} \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

Similarly

$\displaystyle \sf{ \frac{ \partial}{ \partial y} \bigg [ y{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] } = \displaystyle \sf{ \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 {y}^{2} \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

And

$\displaystyle \sf{ \frac{ \partial}{ \partial z} \bigg [ z{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] } = \displaystyle \sf{ \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 {z}^{2} \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

So

$\displaystyle \sf{ \nabla \bigg( \frac{ \vec{r}}{ {r}^{3} } \bigg) }$

$= \displaystyle \sf{ \frac{ \partial}{ \partial x} \bigg [ x{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] + \frac{ \partial}{ \partial y} \bigg [ y{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] \: +\frac{ \partial}{ \partial z} \bigg [ z{({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] \: }$

$= \displaystyle \sf{ 3\bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 ({x}^{2} + {y}^{2} + {z}^{2} )\bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{5}{2} } \bigg ] \: }$

$= \displaystyle \sf{ 3\bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] - 3 \bigg [ {({x}^{2} + {y}^{2} + {z}^{2})}^{ - \frac{3}{2} } \bigg ] \: }$

$= \sf{0}$

So in everywhere

$\displaystyle \sf{ \nabla \bigg( \frac{ \vec{r}}{ {r}^{3} } \bigg) } = 0$

RESULT

The required answer is - ( b) ZERO EVERYWHERE