Physics - Elecrostatics Question with Solution | TestHub

Potential for uniformly charged solid sphere

$\text{v}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{r}}$outside i.e r > R

$\text{v}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}$on the surface

$\text{v}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}\left[\frac{3}{2}-\frac{1}{2}\frac{{\text{r}}^2}{{\text{R}}^2}\right]$inside i.e. r < R

Clearly

$\frac{3{\text{v}}_0}{2}\text{,}\frac{5{\text{v}}_0}{4}$are inside potentials$\left[\because >{\text{v}}_0\right]$

$\frac{3{\text{v}}_0}{4}\text{,}\frac{{\text{v}}_0}{4}$are outside potentials$\left[\because <{\text{v}}_0\right]$

To get${\text{R}}_1$:$\frac{3{\text{v}}_0}{2}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}\left[\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_1^2}{{\text{R}}^2}\right]$

${\text{where, v}}_0=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}$

$\frac{3}{2\times 4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}\left[\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_1^2}{{\text{R}}^2}\right]$

$\frac{3}{2}=\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_1^2}{{\text{R}}^2}\Rightarrow {\text{R}}_1=0$

To get${\text{R}}_2$:$\frac{5}{4}{\text{v}}_0=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}\left[\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_2^2}{{\text{R}}^2}\right]$

$\frac{5}{4}\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}\left[\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_2^2}{{\text{R}}^2}\right]$

$\frac{5}{4}=\frac{3}{2}-\frac{1}{2}\frac{{\text{R}}_2^2}{{\text{R}}^2}$

$\frac{1}{2}\frac{{\text{R}}_2^2}{{\text{R}}^2}=\frac{1}{4 }$$\Rightarrow$${\text{R}}_2=\frac{\text{R}}{\sqrt{2}}$

To get${\text{R}}_3$:$\frac{3{\text{v}}_0}{4}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{{\text{R}}_3}$

$\frac{3}{4}\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{{\text{R}}_3}$

$\frac{3}{4\text{R}}=\frac{1}{{\text{R}}_3}$

${\text{R}}_3=\frac{4}{3}\text{R}$

To get${\text{R}}_4$:$\frac{{\text{v}}_0}{4}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{{\text{R}}_4}$

$\frac{1}{4}\times \frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{\text{R}}=\frac{1}{4\pi {\epsilon }_0}\frac{\text{Q}}{{\text{R}}_4}$

${\text{R}}_4=4\text{R}$

${\text{R}}_4-{\text{R}}_3=4\text{R}-\frac{4\text{R}}{3}=\frac{8\text{R}}{3}>{\text{R}}_2$

${\text{R}}_1=0$and${\text{R}}_2<\left({\text{R}}_4-{\text{R}}_3\right)$