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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2024
  • Hard
  • Numerical

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 _______.

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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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