JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
At the centre of a half ring of radius \(R=10 \mathrm{~cm}\) and linear charge density \(4 \mathrm{n} \mathrm{C} \mathrm{m}^{-1}\), the potential is \(x \pi V\). The value of \(x\) is _______.
- A \(35\)
- B \(36\)
- C \(37\)
- D \(38\)
Answer & Solution
Correct Answer
(B) \(36\)
Step-by-step Solution
Detailed explanation
Potential at centre of half ring \(\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{R}}\) \(\mathrm{V}=\frac{\mathrm{K} \lambda \pi \mathrm{R}}{\mathrm{R}}\) \(\mathrm{V}=\mathrm{K} \lambda \pi \Rightarrow \mathrm{V}=9 \times 10^9 \times 4 \times 10^{-9} \pi\) \(\mathrm{V}=36 \pi\)
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