A) \[{{R}_{1}}=0\]and\[{{R}_{2}}<\left( {{R}_{4}}-{{R}_{3}} \right)\]
B) \[2R<{{R}_{4}}\]
C) \[{{R}_{1}}=0\]and\[{{R}_{2}}>\left( {{R}_{4}}-{{R}_{3}} \right)\]
D) \[{{R}_{1}}\ne 0\]and\[\left( {{R}_{2}}-{{R}_{1}} \right)>\left( {{R}_{4}}-{{R}_{3}} \right)\]
Correct Answer: A
Solution :
| [a] Assuming sphere carries positive charge, |
| Potential on the surface is\[{{V}_{0}}=\frac{KQ}{R}\] |
| Potential expression inside the sphere is \[{{V}_{inside}}=\frac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{r}^{2}} \right)\] |
| Potential at the centre is |
| \[{{V}_{centre}}=\frac{3KQ}{2{{R}^{3}}}\left( \because r=0 \right)=\frac{3{{V}_{0}}}{2}\] |
| So radius of equi-potential at the centre of sphere is zero. \[{{R}_{1}}=0\] |
| Let radius of 2nd equi - potential surface is \[{{R}_{2}}\] then |
| \[\frac{5{{V}_{0}}}{4}=\frac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-R_{2}^{2} \right)\]\[\frac{5{{V}_{0}}}{182}=\frac{{{V}_{0}}}{2{{R}^{3}}}\left( 3{{R}^{2}}-R_{2}^{2} \right)\] |
| \[5{{R}^{2}}=6{{R}^{2}}-2R_{2}^{2}\]\[2R_{2}^{2}={{R}^{2}}\Rightarrow {{R}_{2}}=\frac{R}{\sqrt{2}}\] |
| Let radius of 3rd equi-potential is \[{{R}_{3}}\]then |
| \[\frac{3{{V}_{0}}}{4}=\frac{KQ}{{{R}_{3}}}\] |
| \[\frac{3KQ}{4R}=\frac{KQ}{{{R}_{3}}}\Rightarrow {{R}_{3}}=\frac{4R}{3}\] |
| Let radius of 4th equi-potential is\[{{R}_{4}}\] then |
| \[\frac{{{V}_{0}}}{4}=\frac{KQ}{{{R}_{4}}}\] |
| \[\frac{KQ}{4R}=\frac{KQ}{{{R}_{4}}}\Rightarrow {{R}_{4}}=4R\] |
| \[{{R}_{4}}-{{R}_{3}}=4R-\frac{4R}{3}=\frac{8R}{3}=2.66R\] |
| \[{{R}_{2}}=\frac{R}{1.414}<R\] so, \[{{R}_{2}}<{{R}_{4}}-{{R}_{3}}\] |
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