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JEE Main 2002 (AIEEE) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2002 (AIEEE)

Q127JEE Main 2002 · AIEEEEasy

Electric Potential & Equipotentials

On moving a charge of 2020 coulomb by 2 cm,2 J2\text{ cm}, 2\text{ J} of work is done, then the potential difference between the points is

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Answer: (a)

  1. W=qV⇒V=WqW = qV \Rightarrow V = \frac{W}{q} (the 2 cm distance is not needed)
  2. V=220=0.1 VV = \frac{2}{20} = 0.1\text{ V}
  3. So the answer is (a) 0.1 V0.1\text{ V}.
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