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JEE Main 2024 (9 Apr, Shift 1) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2024 (9 Apr, Shift 1)

Q52JEE Main 2024 · 9 APR · SHIFT 1Medium

Electric Potential & Equipotentials

At the centre of a half ring of radius R=10 cmR = 10\text{ cm} and linear charge density 4 nC m−14\text{ nC m}^{-1}, the potential is xπ Vx\pi\text{ V}. The value of xx is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 36

  1. Half ring: V=k(λπR)R=kλπV = \dfrac{k(\lambda\pi R)}{R} = k\lambda\pi
  2. V=9×109×4×10−9×π=36πV = 9\times10^9\times4\times10^{-9}\times\pi = 36\pi V
  3. So x=36x = 36.
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