JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A uniformly charged solid sphere of radius \(R\) has potential \(V_0\) (measured with respect to \(\infty\)) on its surface. For this sphere the equipotential surfaces with potentials \(\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}\) and \(\frac{{{V_0}}}{4}\) have rasius \(R_1,R_2,R_3\) and \(R_4\) respectively. Then
- A \(R_1\)\( \ne 0\) and \((R_2-R_1) > (R_4-R_3)\)
- B \(R_1\) \( = 0\) and \(R_2 < (R_4-R_3)\)
- C \(2R < R_4\)
- D \(R_1\) \( = 0\) and \( R_2 > (R_4-R_3)\)
Answer & Solution
Correct Answer
(C) \(2R < R_4\)
Step-by-step Solution
Detailed explanation
We know, \(\mathrm{V}_{0}=\frac{\mathrm{Kq}}{\mathrm{R}}=\mathrm{V\,surface}\) Now, \(\mathrm{V}_{\mathrm{i}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right) \quad[\mathrm{For}\, \mathrm{r}<\mathrm{R}]\) At the centre of sphare \(r=0 .\) Here…
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