A solid sphere has a charge $Q$ distributed in its volume with a charge density $d=kr^a$ where $k$ and $a$ are constants and $r$ is distance from center. If electric field at $r=\frac R2$ is $\frac 18$ times stronger than that of at $r=R$ , find $a$ .

The answer is 2.

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1st part: By Gauss's Law: $\oint { \vec { E } \cdot \vec { dA } } =\frac { Q }{ \varepsilon }$

At the point r=R/2: ${ E }_{ 1 }\pi \left( \frac { R }{ 2 } \right) ^{ 2 }=\frac { { Q }_{ 1 } }{ \varepsilon }$

At the point r=R: ${ E }_{ 2 }\pi { R }^{ 2 }=\frac { { Q }_{ 2 } }{ \varepsilon }$

We know that E2=8*E1, dividing the equations we get that: $\boxed{32Q1=Q2}$

2nd part (to avoid confusion, density will be "D", and derivatives "d"): $D=K{ r }^{ a }=\frac { dQ }{ dV } \rightarrow \int { dQ } =K\int { { r }^{ a }dV }$

$V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\rightarrow \frac { dV }{ dr } =4\pi { r }^{ 2 }$

$\int { dQ } =K4\pi \int { { r }^{ a+2 }dr } \rightarrow Q=\frac { 4\pi K }{ a+3 } { r }^{ a+3 }$

$32{ Q }_{ 1 }={ Q }_{ 2 }\rightarrow 32\frac { 4\pi K }{ a+3 } \frac { { R }^{ a+3 } }{ { 2 }^{ a+3 } } =\frac { 4\pi K }{ a+3 } { { R }^{ a+3 } }\rightarrow 32={ 2 }^{ a+3 }$

Therefore, $\boxed{a=2}$

I don't know if my solution is the easiest way of doing it. I think it's not considering this is a level 2 problem and my solution is not so simple. If you have a simpler solution please post it.