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Electric Charges and Fields — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields3 Marks
Question
Consider a sphere of radius R with charge density distributed as$\rho\text{(r)}=\text{kr for r}\leq\text{R}$
$=0\text{ for f}>\text{R}.$
Find the electric field at all points r.
✓

Answer

The expression of charge density distribution in the sphere suggests that the electric field is radial. Let us consider a sphere S of radius R and two hypothetic spheres of radius r < R and r > R. Let us first consider for point r < R, electric field intensity will be given by, $\oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\in_0}\int\rho\text{dV}$
Here $\text{dV}=4\pi\text{r}^2\text{dr}$$\Rightarrow\ \oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\in_0}4\pi\text{K}\int\text{r}^3\text{dr}\ \ (\because\ \rho(\text{r})=\text{Kr})$
$\Rightarrow\ (\text{E})4\pi\text{r}^2=\frac{4\pi\text{K}}{\in_0}\frac{\text{r}^4}{4}$
We get, $\text{E}=\frac{1}{4\in_0}\text{Kr}^2$ As charge density is positive, it means the direction of E is radially outwards. Now consider points r > R, electric field intensity will be given by$\oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\in_0}\int\rho\text{dV}$
$\Rightarrow\ \text{E}(4\pi\text{r}^2)=\frac{4\pi\text{K}}{\in_0}\oint\text{r}^3\text{dr}=\frac{4\pi\text{K}}{\in_0}\frac{\text{R}^4}{4}$
Which given, $\text{E}=\frac{\text{K}}{4\in_0}\frac{\text{r}^4}{\text{r}^2}$ Here also the charge density is again positive. So, the direction of E is radially outward.

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