A charged sphere of radius R carries a positive charge whose volume charge density depend only on the distancer from the ball's centre as ρ=ρ01-rR, where ρ0 is a constant. Assume the permittivity of space ∈.The magnitude of electric field as a function of the distance r inside the sphere is given by

According to Gauss's theorem,E=14πε0qin r2For r≤Rqin=∫0r4πr2·dr·ρ =∫0r4πr2ρ01-rR dr =4πρ0r33-r44RSubstituting in Eq. (i), we getE=ρ0εr3-r24R

A charged sphere of radius R carries a positive charge whose volume charge density depend only on the distancer from the ball's centre as ρ=ρ01-rR, where ρ0 is a constant. Assume the permittivity of space ∈.The magnitude of electric field as a function of the distance r inside the sphere is given by

According to Gauss's theorem,E=14πε0qin r2For r≤Rqin=∫0r4πr2·dr·ρ =∫0r4πr2ρ01-rR dr =4πρ0r33-r44RSubstituting in Eq. (i), we getE=ρ0εr3-r24R

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A charged sphere of radius $R$ carries a positive charge whose volume charge density depend only on the distancer from the ball's centre as $ρ = ρ_{0} (1 - \frac{r}{R})$ , where $ρ_{0}$ is a constant. Assume the permittivity of space $\in .$
The magnitude of electric field as a function of the distance $r$ inside the sphere is given by

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$E = \frac{ρ_{0}}{ε} [\frac{r}{3} - \frac{r^{2}}{4 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{4} - \frac{r^{2}}{3 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{3} + \frac{r^{2}}{4 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{4} + \frac{r^{2}}{3 R}]$

answer is A.

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Detailed Solution

According to Gauss's theorem,

$E = (\frac{1}{4 π ε_{0}}) (\frac{q_{in}}{r^{2}})$

For $r \leq R$

$q_{i n} = \int_{0}^{r} (4 π r^{2}) \cdot d r \cdot ρ = \int_{0}^{r} (4 π r^{2}) (ρ_{0}) (1 - \frac{r}{R}) d r = 4 π ρ_{0} (\frac{r^{3}}{3} - \frac{r^{4}}{4 R})$

Substituting in Eq. (i), we get

$E = \frac{ρ_{0}}{ε} [\frac{r}{3} - \frac{r^{2}}{4 R}]$

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