A solid ball of radius $R$ has a charge density $ρ$ given by $ρ = ρ_{0} (1 - r / R)$ for $0 \leq r \leq R$ . The electric field outside the ball is:

Question

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Answer 1

$ρ = ρ_{0} (1 - \frac{r}{R})$
$d q = ρ d v$
$q_{i n} = \int d q$
$= ρ d v$
$ρ_{0} (1 - \frac{r}{R}) 4 π r^{2} d r$

$∵ d v = 4 π r^{2} d r$

$= 4 π ρ_{0} \int_{0}^{R} (1 - \frac{r}{R}) r^{2} d r$

$= 4 π ρ_{o} \int_{o}^{R} r^{2} d r - \frac{r ^{2}}{R} d r$

$= 4 π ρ_{o} [[\frac{r ^{3}}{3}]_{o}^{R} - [\frac{r ^{4}}{4 R}]_{o}^{R}]$

$= 4 π ρ_{o} [\frac{R ^{3}}{3} - \frac{R ^{4}}{4 R}]$

$= 4 π ρ_{o} [\frac{R ^{3}}{3} - \frac{R ^{3}}{4}]$

$= 4 π ρ_{o} [\frac{R ^{3}}{1 2}]$

$q = \frac{π ρ _{o} R ^{3}}{3}$

$E . 4 π r^{2} = (\frac{π ρ _{o} R ^{3}}{3 \in _{o}})$

$\Rightarrow E = \frac{ρ _{o} R ^{3}}{1 2 \in _{o} r ^{2}}$

Answer 2

ρ = ρ_{0} (1 - \frac{r}{R})

d q = ρ d v

q_{i n} = \int d q

= ρ d v

ρ_{0} (1 - \frac{r}{R}) 4 π r^{2} d r

∵ d v = 4 π r^{2} d r

= 4 π ρ_{0} \int_{0}^{R} (1 - \frac{r}{R}) r^{2} d r

= 4 π ρ_{o} \int_{o}^{R} r^{2} d r - \frac{r ^{2}}{R} d r

= 4 π ρ_{o} [[\frac{r ^{3}}{3}]_{o}^{R} - [\frac{r ^{4}}{4 R}]_{o}^{R}]

= 4 π ρ_{o} [\frac{R ^{3}}{3} - \frac{R ^{4}}{4 R}]

= 4 π ρ_{o} [\frac{R ^{3}}{3} - \frac{R ^{3}}{4}]

= 4 π ρ_{o} [\frac{R ^{3}}{1 2}]

q = \frac{π ρ _{o} R ^{3}}{3}

E . 4 π r^{2} = (\frac{π ρ _{o} R ^{3}}{3 \in _{o}})

\Rightarrow E = \frac{ρ _{o} R ^{3}}{1 2 \in _{o} r ^{2}}