JEE Main 2015 — Physics Electromagnetism

Potential for uniformly charged solid sphere

$\text{v} = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{r} }$v=4πϵ0​1​rQ​ outside i.e r > R

$\text{v} = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} }$v=4πϵ0​1​RQ​ on the surface

$\text{v} = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } [ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{r}}^{2} }{ {\text{R}}^{2} } ]$v=4πϵ0​1​RQ​[23​−21​R2r2​] inside i.e. r < R

Clearly

$\frac{ 3 {\text{v}}_{0} }{ 2 } \text{,} \frac{ 5 {\text{v}}_{0} }{ 4 }$23v0​​,45v0​​ are inside potentials $[ \because > {\text{v}}_{0} ]$[∵>v0​]

$\frac{ 3 {\text{v}}_{0} }{ 4 } \text{,} \frac{ {\text{v}}_{0} }{ 4 }$43v0​​,4v0​​ are outside potentials $[ \because < {\text{v}}_{0} ]$[∵<v0​]

To get ${\text{R}}_{1}$R1​: $\frac{ 3 {\text{v}}_{0} }{ 2 } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } [ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{1}^{2} }{ {\text{R}}^{2} } ]$23v0​​=4πϵ0​1​RQ​[23​−21​R2R12​​]

$${\text{where, v}}_{0} = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} }$$where, v0​=4πϵ0​1​RQ​

$$\frac{ 3 }{ 2 \times 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } [ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{1}^{2} }{ {\text{R}}^{2} } ]$$2×4πϵ0​3​RQ​=4πϵ0​1​RQ​[23​−21​R2R12​​]

$$\frac{ 3 }{ 2 } = \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{1}^{2} }{ {\text{R}}^{2} } \Rightarrow {\text{R}}_{1} = 0$$23​=23​−21​R2R12​​⇒R1​=0

To get ${\text{R}}_{2}$R2​: $\frac{ 5 }{ 4 } {\text{v}}_{0} = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } [ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{2}^{2} }{ {\text{R}}^{2} } ]$45​v0​=4πϵ0​1​RQ​[23​−21​R2R22​​]

$$\frac{ 5 }{ 4 } \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } [ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{2}^{2} }{ {\text{R}}^{2} } ]$$45​4πϵ0​1​RQ​=4πϵ0​1​RQ​[23​−21​R2R22​​]

$$\frac{ 5 }{ 4 } = \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } \frac{ {\text{R}}_{2}^{2} }{ {\text{R}}^{2} }$$45​=23​−21​R2R22​​

$\frac{ 1 }{ 2 } \frac{ {\text{R}}_{2}^{2} }{ {\text{R}}^{2} } = \frac{ 1 }{ 4 }$21​R2R22​​=41​$\Rightarrow$⇒ ${\text{R}}_{2} = \frac{ \text{R} }{ \sqrt{ 2 } }$R2​=2​R​

To get ${\text{R}}_{3}$R3​ : $\frac{ 3 {\text{v}}_{0} }{ 4 } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ {\text{R}}_{3} }$43v0​​=4πϵ0​1​R3​Q​

$$\frac{ 3 }{ 4 } \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ {\text{R}}_{3} }$$43​4πϵ0​1​RQ​=4πϵ0​1​R3​Q​

$$\frac{ 3 }{ 4 \text{R} } = \frac{ 1 }{ {\text{R}}_{3} }$$4R3​=R3​1​

$${\text{R}}_{3} = \frac{ 4 }{ 3 } \text{R}$$R3​=34​R

To get ${\text{R}}_{4}$R4​: $\frac{ {\text{v}}_{0} }{ 4 } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ {\text{R}}_{4} }$4v0​​=4πϵ0​1​R4​Q​

$$\frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ \text{R} } = \frac{ 1 }{ 4 \pi {\epsilon }_{0} } \frac{ \text{Q} }{ {\text{R}}_{4} }$$41​×4πϵ0​1​RQ​=4πϵ0​1​R4​Q​

$${\text{R}}_{4} = 4 \text{R}$$R4​=4R

$${\text{R}}_{4} - {\text{R}}_{3} = 4 \text{R} - \frac{ 4 \text{R} }{ 3 } = \frac{ 8 \text{R} }{ 3 } > {\text{R}}_{2}$$R4​−R3​=4R−34R​=38R​>R2​

${\text{R}}_{1} = 0$R1​=0 and ${\text{R}}_{2} < ( {\text{R}}_{4} - {\text{R}}_{3} )$R2​<(R4​−R3​)