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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A point charge of magnitude \(+ 1\,\mu C\) is fixed at \((0, 0, 0) \). An isolated uncharged spherical conductor, is fixed with its center at \((4, 0, 0).\) The potential and the induced electric field at the centre of the sphere is
- A \(1.8\times 10^5\,V\) and \(- 5.625 \times 10^6\,V/m\)
- B \(0\,V\) and \(0\,V/m\)
- C \(2.25 \times 10^5\,V\) and \(-5.625 \times 10^6\,V/m\)
- D \(2.25 \times 10^5\,V\) and \(0\,V/m\)
Answer & Solution
Correct Answer
(C) \(2.25 \times 10^5\,V\) and \(-5.625 \times 10^6\,V/m\)
Step-by-step Solution
Detailed explanation
\(\mathrm{q}=1\, \mu \mathrm{C}=1 \times 10^{-6} \,\mathrm{C}\) \(r=4\, \mathrm{cm}=4 \times 10^{-2}\, \mathrm{m}\) \( \text { Potential } V =\frac{\mathrm{k} q}{\mathrm{r}} \) \(=\frac{9 \times 10^{9} \times 10^{-6}}{4 \times 10^{-2}} \) \(=2.25 \times 10^{5}\, \mathrm{V} \)…
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