JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
The two thin coaxial rings, each of radius \('a'\) and having charges \(+{Q}\) and \(-{Q}\) respectively are separated by a distance of \('s'.\) The potential difference between the centres of the two rings is :
- A \(\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)
- B \(\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)
- C \(\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)
- D \(\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)
Answer & Solution
Correct Answer
(D) \(\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)
Step-by-step Solution
Detailed explanation
\({V}_{{A}}=\frac{{KQ}}{{a}}-\frac{{KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\) \({V}_{{B}}=\frac{-{KQ}}{{a}}+\frac{{KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\) \({V}_{{A}}-{V}_{{B}}=\frac{2 {KQ}}{{a}}-\frac{2 {KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\)…
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