A solid ball of radius R has a charge density ρ given by ρ=ρ_o1-r/R for 0≤r≤R. The electric ﬁeld outside the ball is

Correct answer: (D) (ρ_0R^3)/(12ε_0r^2). ρ=ρ_01-r/R dq=ρdv q_in=∫dq=ρdv ⇒q_in=∫0Rρ_01-r/R4πr^2dr =4πρ_0∫0R1-r/Rr^2dr =4πρ_0∫0Rr^2dr-r^3Rdr =4πρ_0r^330R-r^44R0R =4πρ_0 (R^3)/3-(R^4)/4R =4πρ_0(R^3)/12 ⇒q_in=(πρ_0R^3)/3 E×4πr^2=(πρ_0R^3)/(3ε_0) ⇒E=(ρ_0R^3)/(12ε_0r^2)

Q8 / 142 · NTA Abhyas (JEE Main) 2020 · Physics · Electrostatics

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

A
$\frac{ρ_{0} R^{3}}{ε_{0} r^{2}}$
B
$\frac{4 ρ_{0} R^{3}}{3 ε_{0} r^{2}}$
C
$\frac{3 ρ_{0} R^{3}}{4 ε_{0} r^{2}}$
Correct answer
D
$\frac{ρ_{0} R^{3}}{12 ε_{0} r^{2}}$

Solution

$ρ = ρ_{0} (1 - \frac{r}{R})$

$d q = ρ d v$

$q_{i n} = \int d q = ρ d v$

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

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

$= 4 π ρ_{0} \int_{0}^{R} r^{2} d r - \frac{r^{3}}{R} d r$

$= 4 π ρ_{0} [{[\frac{r^{3}}{3}]}_{0}^{R} - {[\frac{r^{4}}{4 R}]}_{0}^{R}]$

$= 4 π ρ_{0} [\frac{R^{3}}{3} - \frac{R^{4}}{4 R}]$

$= 4 π ρ_{0} [\frac{R^{3}}{12}]$

$\Rightarrow q_{in} = \frac{π ρ_{0} R^{3}}{3}$

$E \times 4 π r^{2} = (\frac{π ρ_{0} R^{3}}{3 ε_{0}})$

$\Rightarrow E = \frac{ρ_{0} R^{3}}{12 ε_{0} r^{2}}$

Hence, the correct answer is option: 'D'.

A solid ball of radius R has a charge density ρ given by ρ=ρo1-rR for 0≤r≤R. The electric ﬁeld outside the ball is

Solution

More Electrostatics Questions

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